= nlog n −n + 1.
As to the number of element assignments, there are, by Observation 1.2
applied to each merge operation, 2n element assignments in each iteration
of the outer while loop for a total of 2nlog n. As a result, we have the
following observation:
Observation 1.5 The total number of element comparisons performed
by Algorithm bottomupsort to sort an array of n elements, where n is a
power of 2, is between (nlog n)/2 and nlog n −n +1. The total number of
element assignments done by the algorithm is exactly 2nlog n.
1.8 Time Complexity
In this section we study an essential component of algorithmic analysis,
namely determining the running time of an algorithm. This theme belongs
to an important area in the theory of computation, namely computational
complexity, which evolved when the need for eﬃcient algorithms arose in
the 1960’s and ﬂourished in the 1970’s and 1980’s. The main objects of
study in the ﬁeld of computational complexity include the time and space
needed by an algorithm in order to deliver its output when presented with
Time Complexity 21
legal input. We start this section with an example whose sole purpose is to
reveal the importance of analyzing the running time of an algorithm.
Example 1.4 We have shown before that the maximum number of element
comparisons performed by Algorithm bottomupsort when n is a power of 2 is
nlog n −n +1, and the number of element comparisons performed by Algorithm
selectionsort is n(n−1)/2. The elements may be integers, real numbers, strings
of characters, etc. For concreteness, let us assume that each element comparison
takes 10
−6
seconds on some computing machine. Suppose we want to sort a
small number of elements, say 128. Then the time taken for comparing elements
using Algorithm bottomupsort is at most 10
−6
(128 7 − 128 + 1) = 0.0008
seconds. Using Algorithm selectionsort, the time becomes 10
−6
(128127)/2 =
0.008 seconds. In other words, Algorithm bottomupsort uses one tenth of the
time taken for comparison using Algorithm selectionsort. This, of course,
is not noticeable, especially to a novice programmer whose main concern is to
come up with a program that does the job. However, if we consider a larger
number, say n = 2
20
= 1, 048, 576 which is typical of many real world problems,
we ﬁnd the following: The time taken for comparing elements using Algorithm
bottomupsort is at most 10
−6
(2
20
20 −2
20
+1) = 20 seconds, whereas, using
Algorithm selectionsort, the time becomes 10
−6
(2
20
(2
20
−1))/2 = 6.4 days!
The calculations in the above example reveal the fact that time is un-
doubtedly an extremely precious resource to be investigated in the analysis
of algorithms.
1.8.1 Order of growth
Obviously, it is meaningless to say that an algorithm A, when presented
with input x, runs in time y seconds. This is because the actual time is
not only a function of the algorithm used: it is a function of numerous
factors, e.g. how and on what machine the algorithm is implemented and
in what language or even what compiler or programmer’s skills, to mention
a few. Therefore, we should be content with only an approximation of the
exact time. But, ﬁrst of all, when assessing an algorithm’s eﬃciency, do
we have to deal with exact or even approximate times? It turns out that
we really do not need even approximate times. This is supported by many
factors, some of which are the following. First, when analyzing the running
time of an algorithm, we usually compare its behavior with another algo-
22 Basic Concepts in Algorithmic Analysis
rithm that solves the same problem, or even a diﬀerent problem. Thus, our
estimates of times are relative as opposed to absolute. Second, it is desir-
able for an algorithm to be not only machine independent, but also capable
of being expressed in any language, including human languages. Moreover,
it should be technology independent, that is, we want our measure of the
running time of an algorithm to survive technological advances. Third, our
main concern is not in small input sizes; we are mostly concerned with the
behavior of the algorithm under investigation on large input instances.
In fact, counting the number of operations in some “reasonable” imple-
mentation of an algorithm is more than what is needed. As a consequence
of the third factor above, we can go a giant step further: A precise count
of the number of all operations is very cumbersome, if not impossible, and
since we are interested in the running time for large input sizes, we may talk
about the rate of growth or the order of growth of the running time. For
instance, if we can come up with some constant c > 0 such that the running
time of an algorithm A when presented with an input of size n is at most
cn
2
, c becomes inconsequential as n gets bigger and bigger. Furthermore,
specifying this constant does not bring about extra insight when comparing
this function with another one of diﬀerent order, say dn
3
for an algorithm
B that solves the same problem. To see this, note that the ratio between
the two functions is dn/c and, consequently, the ratio d/c has virtually no
eﬀect as n becomes very large. The same reasoning applies to lower order
terms as in the function f(n) = n
2
log n +10n
2
+n. Here, we observe that
the larger the value of n the lesser the signiﬁcance of the contribution of the
lower order terms 10n
2
and n. Therefore, we may say about the running
times of algorithms A and B above to be “of order” or “in the order of ”
n
2
and n
3
, respectively. Similarly, we say that the function f(n) above is
of order n
2
log n.
Once we dispose of lower order terms and leading constants from a
function that expresses the running time of an algorithm, we say that we
are measuring the asymptotic running time of the algorithm. Equivalently,
in the analysis of algorithms terminology, we may refer to this asymptotic
time using the more technical term “time complexity”.
Now, suppose that we have two algorithms A
1
and A
2
of running times
in the order of nlog n. Which one should we consider to be preferable
to the other? Technically, since they have the same time complexity, we
say that they have the same running time within a multiplicative constant,
that is, the ratio between the two running times is constant. In some
Time Complexity 23
cases, the constant may be important and more detailed analysis of the
algorithm or conducting some experiments on the behavior of the algorithm
may be helpful. Also, in this case, it may be necessary to investigate other
factors, e.g. space requirements and input distribution. The latter is helpful
in analyzing the behavior of an algorithm on the average.
2 3 4 5 6 7 8 9 10
10
20
30
40
50
60
n
n log n
(1,0)
input size
r
u
n
n
i
n
g

is one microsecond and “cent” stands for centuries.
Deﬁnition 1.1 We denote by an “elementary operation” any computa-
tional step whose cost is always upperbounded by a constant amount of
time regardless of the input data or the algorithm used.
Let us take, for instance, the operation of adding two integers. For the
running time of this operation to be constant, we stipulate that the size
of its operands be ﬁxed no matter what algorithm is used. Furthermore,
as we are now dealing with the asymptotic running time, we can freely
choose any positive integer k to be the “word length” of our “model of
computation”. Incidentally, this is but one instance in which the beauty of
asymptotic notation shows oﬀ; the word length can be any ﬁxed positive
integer. If we want to add arbitrarily large numbers, an algorithm whose
running time is proportional to its input size can easily be written in terms
of the elementary operation of addition. Likewise, we can choose from a
Time Complexity 25
large pool of operations and apply the ﬁxed-size condition to obtain as many
number of elementary operations as we wish. The following operations on
ﬁxed-size operands are examples of elementary operation.
• Arithmetic operations: addition, subtraction, multiplication and
division.
• Comparisons and logical operations.
• Assignments, including assignments of pointers when, say, travers-
ing a list or a tree.
In order to formalize the notions of order of growth and time complexity,
special mathematical notations have been widely used. These notations
make it convenient to compare and analyze running times with minimal
use of mathematics and cumbersome calculations.
1.8.2 The O-notation
We have seen before (Observation 1.4) that the number of elementary op-
erations performed by Algorithm insertionsort is at most cn
2
, where c
is some appropriately chosen positive constant. In this case, we say that
the running time of Algorithm insertionsort is O(n
2
) (read “Oh of n
2
”
or “big-Oh of n
2
”). This can be interpreted as follows. Whenever the num-
ber of elements to be sorted is equal to or exceeds some threshold n
0
, the
running time is at most cn
2
for some constant c > 0. It should be empha-
sized, however, that this does not mean that the running time is always as
large as cn
2
, even for large input sizes. Thus, the O-notation provides an
upper bound on the running time; it may not be indicative of the actual
running time of an algorithm. For example, for any value of n, the running
time of Algorithm insertionsort is O(n) if the input is already sorted in
nondecreasing order.
In general, we say that the running time of an algorithm is O(g(n)), if
whenever the input size is equal to or exceeds some threshold n
0
, its running
time can be bounded above by some positive constant c times g(n). The
formal deﬁnition of this notation is as follows.
‡
Deﬁnition 1.2 Let f(n) and g(n) be two functions from the set of natural
‡
The more formal deﬁnition of this and subsequent notations is in terms of sets. We prefer
not to use their exact formal deﬁnitions, as it only complicates things unnecessarily.
26 Basic Concepts in Algorithmic Analysis
numbers to the set of nonnegative real numbers. f(n) is said to be O(g(n))
if there exists a natural number n
0
and a constant c > 0 such that
∀ n ≥ n
0
, f(n) ≤ cg(n).
Consequently, if lim
n→∞
f(n)/g(n) exists, then
lim
n→∞
f(n)
g(n)
= ∞ implies f(n) = O(g(n)).
Informally, this deﬁnition says that f grows no faster than some constant
times g. The O-notation can also be used in equations as a simpliﬁcation
tool. For instance, instead of writing
f(n) = 5n
3
+ 7n
2
−2n + 13,
we may write
f(n) = 5n
3
+O(n
2
).
This is helpful if we are not interested in the details of the lower order
terms.
1.8.3 The Ω-notation
While the O-notation gives an upper bound, the Ω-notation, on the other
hand, provides a lower bound within a constant factor of the running time.
We have seen before (Observation 1.4) that the number of elementary op-
erations performed by Algorithm insertionsort is at least cn, where c is
some appropriately chosen positive constant. In this case, we say that the
running time of Algorithm insertionsort is Ω(n) (read “omega of n”, or
“big-omega of n”). This can be interpreted as follows. Whenever the num-
ber of elements to be sorted is equal to or exceeds some threshold n
0
, the
running time is at least cn for some constant c > 0. As in the O-notation,
this does not mean that the running time is always as small as cn. Thus,
the Ω-notation provides a lower bound on the running time; it may not be
indicative of the actual running time of an algorithm. For example, for any
value of n, the running time of Algorithm insertionsort is Ω(n
2
) if the
input consists of distinct elements that are sorted in decreasing order.
Time Complexity 27
In general, we say that an algorithm is Ω(g(n)), if whenever the input
size is equal to or exceeds some threshold n
0
, its running time can be
bounded below by some positive constant c times g(n).
This notation is widely used to express lower bounds on problems as
well. In other words, it is commonly used to state a lower bound for any
algorithm that solves a speciﬁc problem. For example, we say that the
problem of matrix multiplication is Ω(n
2
). This is shorthand for saying
“any algorithm for multiplying two n n matrices is Ω(n
2
)”. Likewise,
we say that the problem of sorting by comparisons is Ω(nlog n), to mean
that no comparison-based sorting algorithm with time complexity that is
asymptotically less than nlog n can ever be devised. Chapter 12 is devoted
entirely to the study of lower bounds of problems. The formal deﬁnition of
this notation is symmetrical to that of the O-notation.
Deﬁnition 1.3 Let f(n) and g(n) be two functions from the set of natural
numbers to the set of nonnegative real numbers. f(n) is said to be Ω(g(n))
if there exists a natural number n
0
and a constant c > 0 such that
∀ n ≥ n
0
, f(n) ≥ cg(n).
Consequently, if lim
n→∞
f(n)/g(n) exists, then
lim
n→∞
f(n)
g(n)
= 0 implies f(n) = Ω(g(n)).
Informally, this deﬁnition says that f grows at least as fast as some
constant times g. It is clear from the deﬁnition that
f(n) is Ω(g(n)) if and only if g(n) is O(f(n)).
1.8.4 The Θ-notation
We have seen before that the number of elementary operations performed by
Algorithm selectionsort is always proportional to n
2
(Observation 1.3).
Since each elementary operation takes a constant amount of time, we say
that the running time of Algorithm selectionsort is Θ(n
2
) (read “theta
of n
2
”). This can be interpreted as follows. There exist two constants c
1
and
c
2
associated with the algorithm with the property that on any input of size
28 Basic Concepts in Algorithmic Analysis
n ≥ n
0
, the running time is between c
1
n
2
and c
2
n
2
. These two constants
encapsulate many factors pertaining to the details of the implementation
of the algorithm and the machine and technology used. As stated earlier,
the details of the implementation include numerous factors such as the
programming language used and the programmer’s skill.
It can also be shown that the running time of Algorithm bottomup-
sort is Θ(nlog n) for any positive integer n.
In general, we say that the running time of an algorithm is of order
Θ(g(n)) if whenever the input size is equal to or exceeds some threshold
n
0
, its running time can be bounded below by c
1
g(n) and above by c
2
g(n),
where 0 < c
1
≤ c
2
. Thus, this notation is used to express the exact order
of an algorithm, which implies an exact bound on its running time. The
formal deﬁnition of this notation is as follows.
Deﬁnition 1.4 Let f(n) and g(n) be two functions from the set of natural
numbers to the set of nonnegative real numbers. f(n) is said to be Θ(g(n))
if there exists a natural number n
0
and two positive constants c
1
and c
2
such that
∀ n ≥ n
0
, c
1
g(n) ≤ f(n) ≤ c
2
g(n).
Consequently, if lim
n→∞
f(n)/g(n) exists, then
lim
n→∞
f(n)
g(n)
= c implies f(n) = Θ(g(n)),
where c is a constant strictly greater than 0.
An important consequence of the above deﬁnition is that
f(n) = Θ(g(n)) if and only if f(n) = O(g(n)) and f(n) = Ω(g(n)).
Unlike the previous two notations, the Θ-notation gives an exact pic-
ture of the rate of growth of the running time of an algorithm. Thus, the
running time of some algorithms as insertionsort cannot be expressed
using this notation, as the running time ranges from linear to quadratic.
On the other hand, the running time of some algorithms like Algorithm se-
lectionsort and Algorithm bottomupsort can be described precisely
using this notation.
It may be helpful to think of O as similar to ≤, Ω as similar to ≥
and Θ as similar to =. We emphasized the phrase “similar to” since one
Time Complexity 29
should be cautious not to confuse the exact relations with the asymptotic
notations. For example 100n = O(n) although 100n ≥ n, n = Ω(100n)
although n ≤ 100n and n = Θ(100n) although n = 100n.
1.8.5 Examples
The above O, Ω and Θ notations are not only used to describe the time
complexity of an algorithm; they are so general that they can be applied to
characterize the asymptotic behavior of any other resource measure, say the
amount of space used by an algorithm. Theoretically, they may be used in
conjunction with any abstract function. For this reason, we will not attach
any measures or meanings with the functions in the examples that follow.
We will assume in these examples that f(n) is a function from the set of
natural numbers to the set of nonnegative real numbers.
Example 1.5 Let f(n) = 10n
2
+ 20n. Then, f(n) = O(n
2
) since for all n ≥
1, f(n) ≤ 30n
2
. f(n) = Ω(n
2
) since for all n ≥ 1, f(n) ≥ n
2
. Also, f(n) = Θ(n
2
)
since for all n ≥ 1, n
2
≤ f(n) ≤ 30n
2
. We can also establish these three relations
using the limits as mentioned above. Since lim
n→∞
(10n
2
+20n)/n
2
= 10, we see
that f(n) = O(n
2
), f(n) = Ω(n
2
) and f(n) = Θ(n
2
).
Example 1.6 In general, let f(n) = a
k
n
k
+a
k−1
n
k−1
+. . . +a
1
n+a
0
. Then,
f(n) = Θ(n
k
). Recall that this implies that f(n) = O(n
k
) and f(n) = Ω(n
k
).
Example 1.7 Since
lim
n→∞
log n
2
n
= lim
n→∞
2 log n
n
= lim
n→∞
2
ln 2
ln n
n
=
2
ln 2
lim
n→∞
1
n
= 0
(diﬀerentiate both numerator and denominator), we see that f(n) is O(n), but
not Ω(n). It follows that f(n) is not Θ(n).
Example 1.8 Since log n
2
= 2 log n, we immediately see that log n
2
=
Θ(log n). In general, for any ﬁxed constant k, log n
k
= Θ(log n).
Example 1.9 Any constant function is O(1), Ω(1) and Θ(1).
Example 1.10 It is easy to see that 2
n
is Θ(2
n+1
). This is an example of
many functions that satisfy f(n) = Θ(f(n + 1)).
Example 1.11 In this example, we give a monotonic increasing function f(n)
such that f(n) is not Ω(f(n + 1)) and hence not Θ(f(n + 1)). Since (n + 1)! =
30 Basic Concepts in Algorithmic Analysis
(n + 1)n! > n!, we have that n! = O((n + 1)!). Since
lim
n→∞
n!
(n + 1)!
= lim
n→∞
1
n + 1
= 0,
we conclude that n! is not Ω((n + 1)!). It follows that n! is not Θ((n + 1)!)
Example 1.12 Consider the series
¸
n
j=1
log j. Clearly,
n
¸
j=1
log j ≤
n
¸
j=1
log n.
That is,
n
¸
j=1
log j = O(nlog n).
Also,
n
¸
j=1
log j ≥
n/2
¸
j=1
log

n
2

= ]n/2 log

n
2

= ]n/2 log n −]n/2.
Thus,
n
¸
j=1
log j = Ω(nlog n).
It follows that
n
¸
j=1
log j = Θ(nlog n).
Example 1.13 We want to ﬁnd an exact bound for the function f(n) =
log n!. First, note that log n! =
¸
n
j=1
log j. We have shown in Example 1.12 that
¸
n
j=1
log j = Θ(nlog n). It follows that log n! = Θ(nlog n).
Example 1.14 Since log n! = Θ(nlog n) and log 2
n
= n, we deduce that
2
n
= O(n!) but n! is not O(2
n
). Similarly, since log 2
n
2
= n
2
> nlog n, and
log n! = Θ(nlog n) (Example 1.13), it follows that n! = O(2
n
2
), but 2
n
2
is not
O(n!).
Example 1.15 It is easy to see that
n
¸
j=1
n
j
≤
n
¸
j=1
n
1
= O(n
2
).
Time Complexity 31
However, this upper bound is not useful since it is not tight. We will show in
Example 2.16 that
log(n + 1)
log e
≤
n
¸
j=1
1
j
≤
log n
log e
+ 1.
That is
n
¸
j=1
1
j
= O(log n) and
n
¸
j=1
1
j
= Ω(log n).
It follows that
n
¸
j=1
n
j
= n
n
¸
j=1
1
j
= Θ(nlog n).
Example 1.16 Consider the brute-force algorithm for primality test given in
Algorithm brute-force primalitytest.
Algorithm 1.7 brute-force primalitytest
Input: A positive integer n ≥ 2.
Output: true if n is prime and false otherwise.
1. s←]
√
n
2. for j ←2 to s
3. if j divides n then return false
4. end for
5. return true
We will assume here that
√
n can be computed in O(1) time. Clearly, the
algorithm is O(
√
n), since the number of iterations is exactly ]
√
n −1 when the
input is prime. Besides, the number of primes is inﬁnite, which means that the
algorithm performs exactly ]
√
n − 1 iterations for an inﬁnite number of values
of n. It is also easy to see that for inﬁnitely many values of n, the algorithm
performs only O(1) iterations (e.g. when n is even), and hence the algorithm is
Ω(1). Since the algorithm may take Ω(
√
n) time on some inputs and O(1) time
on some other inputs inﬁnitely often, it is neither Θ(
√
n) nor Θ(1). It follows
that the algorithm is not Θ(f(n)) for any function f.
1.8.6 Complexity Classes and the o-notation
Let R be the relation on the set of complexity functions deﬁned by f R g if
and only if f(n)= Θ(g(n)). It is easy to see that R is reﬂexive, symmetric
32 Basic Concepts in Algorithmic Analysis
and transitive, i.e., an equivalence relation (see Sec. 2.1.2.1). The equiv-
alence classes induced by this relation are called complexity classes. The
complexity class to which a complexity function g(n) belongs includes all
functions f(n) of order Θ(g(n)). For example, all polynomials of degree 2
belong to the same complexity class n
2
. To show that two functions be-
long to diﬀerent classes, it is useful to use the o-notation(read “little oh”)
deﬁned as follows.
Deﬁnition 1.5 Let f(n) and g(n) be two functions from the set of natural
numbers to the set of nonnegative real numbers. f(n) is said to be o(g(n))
if for every constant c > 0 there exists a positive integer n
0
such that
f(n) < cg(n) for all n ≥ n
0
. Consequently, if lim
n→∞
f(n)/g(n) exists,
then
lim
n→∞
f(n)
g(n)
= 0 implies f(n) = o(g(n)).
Informally, this deﬁnition says that f(n) becomes insigniﬁcant relative
to g(n) as n approaches inﬁnity. It follows from the deﬁnition that
f(n) = o(g(n)) if and only if f(n) = O(g(n)), but g(n) = O(f(n)).
For example, nlog n is o(n
2
) is equivalent to saying that nlog n is O(n
2
)
but n
2
is not O(nlog n).
We also write f(n) ≺ g(n) to denote that f(n) is o(g(n)). Using this
notation, we can concisely express the following hierarchy of complexity
classes.
1 ≺ log log n ≺ log n ≺
√
n ≺ n
3/4
≺ n ≺ nlog n ≺ n
2
≺ 2
n
≺ n! ≺ 2
n
2
.
1.9 Space Complexity
We deﬁne the space used by an algorithm to be the number of memory cells
(or words) needed to carry out the computational steps required to solve
an instance of the problem excluding the space allocated to hold the input.
In other words, it is only the work space required by the algorithm. The
reason for not including the input size is basically to distinguish between
Space Complexity 33
algorithms that use “less than” linear space throughout their computation.
All deﬁnitions of order of growth and asymptotic bounds pertaining to
time complexity carry over to space complexity. It is clear that the work
space cannot exceed the running time of an algorithm, as writing into each
memory cell requires at least a constant amount of time. Thus, if we let
T(n) and S(n) denote, respectively, the time and space complexities of an
algorithm, then S(n) = O(T(n)).
To appreciate the importance of space complexity, suppose we want to
sort n = 2
20
= 1, 048, 576 elements. If we use Algorithm selectionsort,
then we need no extra storage. On the other hand, if we use Algorithm
bottomupsort, then we need n = 1, 048, 576 extra memory cells as a
temporary storage for the input elements (see Example 1.19 below).
In the following examples, we will look at some of the algorithms we
have discussed so far and analyze their space requirements.
Example 1.17 In Algorithm linearsearch, only one memory cell is used
to hold the result of the search. If we add local variables, e.g. for looping, we
conclude that the amount of space needed is Θ(1). This is also the case in
algorithms binarysearch, selectionsort and insertionsort.
Example 1.18 In Algorithm merge for merging two sorted arrays, we need
an auxiliary amount of storage whose size is exactly that of the input, namely n
(Recall that n is the size of the array A[p..r]). Consequently, its space complexity
is Θ(n).
Example 1.19 When attempting to compute an estimate of the space re-
quired by Algorithm bottomupsort, one may ﬁnd it to be complex at ﬁrst.
Nevertheless, it is not diﬃcult to see that the space needed is no more than n,
the size of the input array. This is because we can set aside an array of size n,
say B[1..n], to be used by Algorithm merge as an auxiliary storage for carry-
ing out the merging process. It follows that the space complexity of Algorithm
bottomupsort is Θ(n).
Example 1.20 In this example, we will “devise” an algorithm that uses
Θ(log n) space. Let us modify the Algorithm binarysearch as follows. After
the search terminates, output a sorted list of all those entries of array A that
have been compared against x. This means that after we test x against A[mid]
in each iteration, we must save A[mid] using an auxiliary array, say B, which can
be sorted later. As the number of comparisons is at most ]log n + 1, it is easy
to see that the size of B should be at most this amount, i.e., O(log n).
34 Basic Concepts in Algorithmic Analysis
Example 1.21 An algorithm that outputs all permutations of a given n char-
acters needs only Θ(n) space. If we want to keep these permutations so that they
can be used in subsequent calculations, then we need at least nn! = Θ((n+1)!)
space.
Naturally, in many problems there is a time-space tradeoﬀ: The more
space we allocate for the algorithm the faster it runs, and vice versa. This,
of course, is within limits: in most of the algorithms that we have discussed
so far, increasing the amount of space does not result in a noticeable speed-
up in the algorithm running time. However, it is almost always the case
that decreasing the amount of work space required by an algorithm results
in a degradation in the algorithm’s speed.
1.10 Optimal Algorithms
In Sec. 12.3.2, we will show that the running time of any algorithm that
sorts an array with n entries using element comparisons must be Ω(nlog n)
in the worst case (see Sec. 1.12). This means that we cannot hope for an
algorithm that runs in time that is asymptotically less than nlog n in the
worst case. For this reason, it is commonplace to call any algorithm that
sorts using element comparisons in time O(nlog n) an optimal algorithm
for the problem of comparison-based sorting. By this deﬁnition, it follows
that Algorithm bottomupsort is optimal. In this case, we also say that
it is optimal within a multiplicative constant to indicate the possibility of
the existence of another sorting algorithm whose running time is a constant
fraction of that of bottomupsort. In general, if we can prove that any
algorithm to solve problem Π must be Ω(f(n)), then we call any algorithm
to solve problem Π in time O(f(n)) an optimal algorithm for problem Π.
Incidentally, this deﬁnition, which is widely used in the literature, does
not take into account the space complexity. The reason is twofold. First,
as we indicated before, time is considered to be more precious than space
so long as the space used is within reasonable limits. Second, most of
the existing optimal algorithms compare to each other in terms of space
complexity in the order of O(n). For example, Algorithm bottomupsort,
which needs Θ(n) of space as auxiliary storage, is called optimal, although
there are other algorithms that sort in O(nlog n) time and O(1) space. For
example, Algorithm heapsort, which will be introduced in Sec. 4.2.3, runs
in time O(nlog n) using only O(1) amount of space.
How to Estimate the Running Time of an Algorithm 35
1.11 How to Estimate the Running Time of an Algorithm
As we discussed before, a bound on the running time of an algorithm, be
it upper, lower or exact, can be estimated to within a constant factor if we
restrict the operations used by the algorithm to those we referred to as ele-
mentary operations. Now it remains to show how to analyze the algorithm
in order to obtain the desired bound. Of course, we can get a precise bound
by summing up all elementary operations. This is undoubtedly ruled out,
as it is cumbersome and quite often impossible. There is, in general, no
mechanical procedure by the help of which one can obtain a “reasonable”
bound on the running time or space usage of the algorithm at hand. More-
over, this task is mostly left to intuition and, in many cases, to ingenuity
too. However, in many algorithms, there are some agreed upon techniques
that give a tight bound with straightforward analysis. In the following, we
discuss some of these techniques using simple examples.
1.11.1 Counting the number of iterations
It is quite often the case that the running time is proportional to the number
of passes through while loops and similar constructs. Thus, it follows that
counting the number of iterations is a good indicative of the running time
of an algorithm. This is the case with many algorithms including those for
searching, sorting, matrix multiplication and so forth.
Example 1.22 Consider Algorithm count1, which consists of two nested
loops and a variable count which counts the number of iterations performed by
the algorithm on input n = 2
k
, for some positive integer k.
Algorithm 1.8 count1
Input: n = 2
k
, for some positive integer k.
Output: count = number of times Step 4 is executed.
1. count←0
2. while n ≥ 1
3. for j ←1 to n
4. count←count + 1
5. end for
6. n←n/2
7. end while
8. return count
36 Basic Concepts in Algorithmic Analysis
The while loop is executed k + 1 times, where k = log n. The for loop is
executed n times, and then n/2, n/4, . . . , 1. Therefore, applying Formula 2.11
(page 78) on this geometric series, the number of times Step 4 is executed is
k
¸
j=0
n
2
j
= n
k
¸
j=0
1
2
j
= n(2 −
1
2
k
) = 2n −1 = Θ(n).
Since the running time is proportional to count, we conclude that it is Θ(n).
Example 1.23 Consider Algorithm count2, which consists of two nested
loops and a variable count which counts the number of iterations performed by
the algorithm on input n, which is a positive integer.
Algorithm 1.9 count2
Input: A positive integer n.
Output: count = number of times Step 5 is executed.
1. count←0
2. for i ←1 to n
3. m←]n/i
4. for j ←1 to m
5. count←count + 1
6. end for
7. end for
8. return count
The inner for loop is executed repeatedly for the following values of n:
n, ]n/2, ]n/3, . . . , ]n/n.
Thus, the total number of times Step 5 is executed is
n
¸
i=1

n
i
¸
.
Since
n
¸
i=1

n
i
−1

≤
n
¸
i=1

n
i
¸
≤
n
¸
i=1
n
i
,
we conclude that Step 5 is executed Θ(nlog n) times (see Examples 1.15 and 2.16).
As the running time is proportional to count, we conclude that it is Θ(nlog n).
Example 1.24 Consider Algorithm count3, which consists of two nested
loops and a variable count which counts the number of iterations performed by
How to Estimate the Running Time of an Algorithm 37
the while loop on input n that is of the form 2
2
k
, for some positive integer k.
For each value of i, the while loop will be executed when
j = 2, 2
2
, 2
4
, . . . , 2
2
k
.
That is, it will be executed when
j = 2
2
0
, 2
2
1
, 2
2
2
, . . . , 2
2
k
.
Thus, the number of iterations performed by the while loop is k+1 = log log n+1
for each iteration of the for loop. It follows that the total number of iterations
performed by the while loop, which is the output of the algorithm, is exactly
n(log log n + 1) = Θ(nlog log n). We conclude that the running time of the
algorithm is Θ(nlog log n).
Algorithm 1.10 count3
Input: n = 2
2
k
, for some positive integer k.
Output: Number of times Step 6 is executed.
1. count←0
2. for i ←1 to n
3. j ←2
4. while j ≤ n
5. j ←j
2
6. count←count + 1
7. end while
8. end for
9. return count
Example 1.25 Let n be a perfect square, i.e., an integer whose square root
is integer. Algorithm psum computes for each perfect square j between 1 and n
the sum
¸
j
i=1
i. (Obviously, this sum can be computed more eﬃciently).
We will assume that
√
n can be computed in O(1) time. We compute the
running time of the algorithm as follows. The outer and inner for loops are
executed k =
√
n and j
2
times, respectively. Hence, the number of iterations
performed by the inner for loop is
k
¸
j=1
j
2
¸
i=1
1 =
k
¸
j=1
j
2
=
k(k + 1)(2k + 1)
6
= Θ(k
3
) = Θ(n
1.5
).
It follows that the running time of the algorithm is Θ(n
1.5
).
38 Basic Concepts in Algorithmic Analysis
Algorithm 1.11 psum
Input: n = k
2
for some integer k.
Output:
¸
j
i=1
i for each perfect square j between 1 and n.
1. k←
√
n
2. for j ←1 to k
3. sum[j] ←0
4. for i ←1 to j
2
5. sum[j] ←sum[j] +i
6. end for
7. end for
8. return sum[1..k]
1.11.2 Counting the frequency of basic operations
In some algorithms, it is cumbersome, or even impossible, to make use
of the previous method in order to come up with a tight estimate of its
running time. Unfortunately, at this point we have not covered good exam-
ples of such algorithms. Good examples that will be covered in subsequent
chapters include the single-source shortest path problem, Prim’s algorithm
for ﬁnding minimum spanning trees, depth-ﬁrst search, computing convex
hulls and others. However, Algorithm merge will serve as a reasonable
candidate. Recall that the function of Algorithm merge is to merge two
sorted arrays into one sorted array. In this algorithm, if we try to apply
the previous method, the analysis becomes lengthy and awkward. Now,
consider the following argument which we have alluded to in Sec. 1.4. Just
prior to Step 15 of the algorithm is executed, array B holds the ﬁnal sorted
list. Thus, for each element x ∈ A, the algorithm executes one element
assignment operation that moves x from A to B. Similarly, in Step 15,
the algorithm executes n element assignment operations in order to copy B
back into A. This implies that the algorithm executes exactly 2n element
assignments (Observation 1.2). On the other hand, there is no other opera-
tion that is executed more than 2n times. For example, at most one element
comparison is needed to move each element from A to B (Observation 1.1).
In general, when analyzing the running time of an algorithm, we may
be able to single out one elementary operation with the property that its
frequency is at least as large as any other operation. Let us call such an
operation a basic operation. We can relax this deﬁnition to include any
operation whose frequency is proportional to the running time.
How to Estimate the Running Time of an Algorithm 39
Deﬁnition 1.6 An elementary operation in an algorithm is called a basic
operation if it is of highest frequency to within a constant factor among all
other elementary operations.
Hence, according to this deﬁnition, the operation of element assignment
is a basic operation in Algorithmmerge and thus is indicative of its running
time. By observation 1.2, the number of element assignments needed to
merge two arrays into one array of size n is exactly 2n. Consequently, its
running time is Θ(n). Note that the operation of element comparison is in
general not a basic operation in Algorithm merge, as there may be only
one element comparison throughout the execution of the algorithm. If,
however, the algorithm is to merge two arrays of approximately the same
size (e.g. (n/2)| and (n/2)|), then we may safely say that it is basic for
that special instance. This happens if, for example, the algorithm is invoked
by Algorithm bottomupsort in which case the two subarrays to be sorted
are of sizes (n/2)| and (n/2)|.
In general, this method consists of identifying one basic operation and
utilizing one of the asymptotic notations to ﬁnd out the order of execution
of this operation. This order will be the order of the running time of the
algorithm. This is indeed the method of choice for a large class of problems.
We list here some candidates of these basic operations:
• When analyzing searching and sorting algorithms, we may choose
the element comparison operation if it is an elementary operation.
• In matrix multiplication algorithms, we select the operation of
scalar multiplication.
• In traversing a linked list, we may select the “operation” of setting
or updating a pointer.
• In graph traversals, we may choose the “action” of visiting a node,
and count the number of nodes visited.
Example 1.26 Using this method, we obtain an exact bound for Algorithm
bottomupsort as follows. First, note that the basic operations in this algorithm
are inherited from Algorithm merge, as the latter is called by the former in each
iteration of the while loop. By the above discussion, we may safely choose the
elementary operation of element comparison as the basic operation. By obser-
vation 1.5, the total number of element comparisons required by the algorithm
when n is a power of 2 is between (nlog n)/2 and nlog n − n + 1. This means
that the number of element comparisons when n is a power of 2 is Ω(nlog n)
40 Basic Concepts in Algorithmic Analysis
and O(nlog n), i.e., Θ(nlog n). It can be shown that this holds even if n is not a
power of 2. Since the operation of element comparison used by the algorithm is
of maximum frequency to within a constant factor, we conclude that the running
time of the algorithm is proportional to the number of comparisons. It follows
that the algorithm runs in time Θ(nlog n).
One should be careful, however, when choosing a basic operation, as
illustrated by the following example.
Example 1.27 Consider the following modiﬁcation to Algorithm insertion-
sort. When trying to insert an element of the array in its proper position, we
will not use linear search; instead, we will use a binary search technique similar
to Algorithm binarysearch. Algorithm binarysearch can easily be modiﬁed
so that it does not return 0 when x is not an entry of array A; instead, it returns
the position of x relative to other entries of the sorted array A. For example,
when Algorithm binarysearch is called with A = 2 3 6 8 9 and x = 7,
it returns 4. Incidentally, this shows that using binary search is not conﬁned to
testing for the membership of an element x in an array A; in many algorithms, it
is used to ﬁnd the position of an element x relative to other elements in a sorted
list. Let Algorithm modbinarysearch be some implementation of this binary
search technique. Thus, modbinarysearch(¦2, 3, 6, 8, 9¦, 7) = 4. The modiﬁed
sorting algorithm is given in Algorithm modinsertionsort.
Algorithm 1.12 modinsertionsort
Input: An array A[1..n] of n elements.
Output: A[1..n] sorted in nondecreasing order.
1. for i ←2 to n
2. x←A[i]
3. k ←modbinarysearch (A[1..i −1], x)
4. for j ←i −1 downto k
5. A[j + 1] ←A[j]
6. end for
7. A[k] ←x
8. end for
The total number of element comparisons are those performed by Algorithm
modbinarysearch. Since this algorithm is called n − 1 times, and since the
maximum number of comparisons performed by the binary search algorithm on
an array of size i − 1 is ]log(i −1) + 1 (Theorem 1.1), it follows that the total
How to Estimate the Running Time of an Algorithm 41
number of comparisons done by Algorithm modinsertionsort is at most
n
¸
i=2
(]log(i −1) + 1) = n −1 +
n−1
¸
i=1
]log i ≤ n −1 +
n−1
¸
i=1
log i = Θ(nlog n).
The last equality follows from Example 1.12 and Eq. 2.18 on page 81. One may be
tempted to conclude, based on the false assumption that the operation of element
comparison is basic, that the overall running time is O(nlog n). However, this
is not the case, as the number of element assignments in Algorithm modinser-
tionsort is exactly that in Algorithm insertionsort when the two algorithms
are run on the same input. This has been shown to be O(n
2
) (Observation 1.4).
We conclude that this algorithm runs in time O(n
2
), and not O(nlog n).
In some algorithms, all elementary operations are not basic. In these
algorithms, it may be the case that the frequency of two or more operations
combined together may turn out to be proportional to the running time of
the algorithm. In this case, we express the running time as a function of
the total number of times these operations are executed. For instance, if
we cannot bound the number of either insertions or deletions, but can come
up with a formula that bounds their total, then we may say something like:
There are at most n insertions and deletions. This method is widely used in
graph and network algorithms. Here we give a simple example that involves
only numbers and the two operations of addition and multiplication. There
are better examples that involve graphs and complex data structures.
Example 1.28 Suppose we are given an array A[1..n] of n integers and a
positive integer k, 1 ≤ k ≤ n, and asked to multiply the ﬁrst k integers in A and
add the rest. An algorithm to do this is sketched below. Observe here that there
are no basic operations, since the running time is proportional to the number of
times both additions and multiplications are performed. Thus, we conclude that
there are n elementary operations: multiplications and additions, which implies
a bound of Θ(n). Note that in this example, we could have counted the number
of iterations to obtain a precise measure of the running time as well. This is
because in each iteration, the algorithm takes a constant amount of time. The
total number of iterations is k + (n −k) = n.
1.11.3 Using recurrence relations
In recursive algorithms, a formula bounding the running time is usually
given in the form of a recurrence relation, that is, a function whose deﬁnition
42 Basic Concepts in Algorithmic Analysis
1. prod←1; sum←0
2. for j ←1 to k
3. prod←prod A[j]
4. end for
5. for j ←k + 1 to n
6. sum←sum+A[j]
7. end for
contains the function itself, e.g. T(n) = 2T(n/2) +n. Finding the solution
of a recurrence relation has been studied well to the extent that the solution
of a recurrence may be obtained mechanically (see Sec. 2.8 for a discussion
on recurrence relations). It may be possible to derive a recurrence that
bounds the number of basic operations in a nonrecursive algorithm. For
example, in Algorithm binarysearch, if we let C(n) be the number of
comparisons performed on an instance of size n, we may express the number
of comparisons done by the algorithm using the recurrence
C(n) ≤

1 if n = 1
C(n/2|) + 1 if n ≥ 2.
The solution to this recurrence reduces to a summation as follows.
C(n) ≤ C(n/2|) + 1
= C(n/2|/2|) + 1 + 1
= C(n/4|) + 1 + 1 (Eq. 2.3, page 71)
.
.
.
= log n| + 1.
That is, C(n) ≤ log n| + 1. It follows that C(n) = O(log n). Since the
operation of element comparison is a basic operation in Algorithm bina-
rysearch, we conclude that its time complexity is O(log n).
1.12 Worst case and average case analysis
Consider the problem of adding two n n matrices A and B of integers.
Clearly, the running time expressed in the number of scalar additions of an
algorithm that computes A + B is the same for any two arbitrary n n
matrices A and B. That is, the running time of the algorithm is insensitive
Worst case and average case analysis 43
to the input values; it is dependent only on its size measured in the number
of entries. This is to be contrasted with an algorithm like insertionsort
whose running time is highly dependent on the input values as well. By
Observation 1.4, the number of element comparisons performed on an input
array of size n lies between n −1 and n(n −1)/2 inclusive. This indicates
that the performance of the algorithm is not only a function of n, but also
a function of the original order of the input elements. The dependence
of the running time of an algorithm on the form of input data, not only
its number, is characteristic of many problems. For instance, the process
of sorting is inherently dependent on the relative order of the data to be
sorted. This does not mean that all sorting algorithms are sensitive to
input data. For instance, the number of element comparisons performed
by Algorithm selectionsort on an array of size n is the same regardless
of the form or order of input values, as the number of comparisons done
by the algorithm is a function of n only. More precisely, the time taken
by a comparison-based algorithm to sort a set of n elements depends on
their relative order. For instance, the number of steps required to sort the
numbers 6, 3, 4, 5, 1, 7, 2 is the same as that for sorting the numbers 60, 30,
40, 50, 10, 70, 20. Obviously, it is impossible to come up with a function
that describes the time complexity of an algorithm based on both input size
and form; the latter, deﬁnitely, has to be suppressed.
Consider again Algorithm insertionsort. Let A[1..n] = ¦1, 2, . . . , n¦,
and consider all n! permutations of the elements in A. Each permuta-
tion corresponds to one possible input. The running time of the algorithm
presumably diﬀers from one permutation to another. Consider three per-
mutations: a in which the elements in A are sorted in decreasing order, c
in which the elements in A are already sorted in increasing order and b in
which the elements are ordered randomly (see Fig. 1.6).
Thus, input a is a representative of the worst case of all inputs of size n,
input c is a representative of the best case of all inputs of size n and input
b is between the two. This gives rise to three methodologies for analyzing
the running time of an algorithm: worst case analysis, average case analysis
and best case analysis. The latter is not used in practice, as it does not
give useful information about the behavior of an algorithm in general.
44 Basic Concepts in Algorithmic Analysis
n
input size
worst case
average case
best case
o
/
c
r
u
n
n
i
n
g

t
i
m
e
n
z
/¬
n
z
/i
Fig. 1.6 Performance of Algorithm insertionsort: worst, average and best cases.
1.12.1 Worst case analysis
In worst case analysis of time complexity we select the maximum cost
among all possible inputs of size n. As stated above, for any positive inte-
ger n, Algorithm insertionsort requires Ω(n
2
) to process some inputs of
size n (e.g. input a in Fig. 1.6). For this reason, we say that the running
time of this algorithm is Ω(n
2
) in the worst case. Since the running time
of the algorithm is O(n
2
), we also say that the running time of the algo-
rithm is O(n
2
) in the worst case. Consequently, we may use the stronger
Θ-notation and say that the running time of the algorithm is Θ(n
2
) in the
worst case. Clearly, using the Θ-notation is preferred, as it gives the exact
behavior of the algorithm in the worst case. In other words, stating that
Algorithm insertionsort has a running time of Θ(n
2
) in the worst case
implies that it is also Ω(n
2
) in the worst case, whereas stating that Algo-
rithm insertionsort runs in O(n
2
) in the worst case does not. Note that
for any value of n there are input instances on which the algorithm spends
no more than O(n) time (e.g. input c in Fig. 1.6).
It turns out that under the worst case assumption, the notions of upper
and lower bounds in many algorithms coincide and, consequently, we may
say that an algorithm runs in time Θ(f(n)) in the worst case. As explained
Worst case and average case analysis 45
above, this is stronger than stating that the algorithm is O(f(n)) in the
worst case. As another example, we have seen before that Algorithm lin-
earsearch is O(n) and Ω(1). In the worst case, this algorithm is both
O(n) and Ω(n), i.e., Θ(n).
One may be tempted, however, to conclude that in the worst case the
notions of upper and lower bounds always coincide. This in fact is not the
case. Consider for example an algorithm whose running time is known to
be O(n
2
) in the worst case. However, it has not been proven that for all
values of n greater than some threshold n
0
there exists an input of size n
on which the algorithm spends Ω(n
2
) time. In this case, we cannot claim
that the algorithm’s running time is Θ(n
2
) in the worst case, even if we
know that the algorithm takes Θ(n
2
) time for inﬁnitely many values of n.
It follows that the algorithm’s running time is not Θ(n
2
) in the worst case.
This is the case in many graph and network algorithms for which only an
upper bound on the number of operations can be proven, and whether this
upper bound is achievable is not clear. The next example gives a concrete
instance of this case.
Example 1.29 Consider for example the procedure shown below whose input
is an element x and a sorted array A of n elements.
1. if n is odd then k← binarysearch(A, x)
2. else k← linearsearch(A, x)
This procedure searches for x in A using binary search if n is odd and linear
search if n is even. Obviously, the running time of this procedure is O(n), since
when n is even, the running time is that of Algorithm linearsearch, which is
O(n). However, the procedure is not Ω(n) in the worst case because there does
not exist a threshold n
0
such that for all n ≥ n
0
there exists some input of size
n that causes the algorithm to take at least cn time for some constant c. We can
only ascertain that the running time is Ω(log n) in the worst case. Note that the
running time being Ω(n) for inﬁnitely many values of n does not mean that the
algorithm’s running time is Ω(n) in the worst case. It follows that, in the worst
case, this procedure is O(n) and Ω(log n), which implies that, in the worst case,
it is not Θ(f(n)) for any function f(n).
46 Basic Concepts in Algorithmic Analysis
1.12.2 Average case analysis
Another interpretation of an algorithm’s time complexity is that of the
average case. Here, the running time is taken to be the average time over
all inputs of size n (see Fig. 1.6). In this method, it is necessary to know
the probabilities of all input occurrences, i.e., it requires prior knowledge
of the input distribution. However, even after relaxing some constraints
including the assumption of a convenient input distribution, e.g. uniform
distribution, the analysis is in many cases complex and lengthy.
Example 1.30 Consider Algorithm linearsearch. To simplify the analysis,
let us assume that A[1..n] contains the numbers 1 through n, which implies that all
elements of A are distinct. Also, we will assume that x is in the array, i.e., x ∈
¦1, 2, . . . , n¦. Furthermore, and most importantly indeed, we will assume that
each element y in A is equally likely to be in any position in the array. In other
words, the probability that y = A[j] is 1/n, for all y ∈ A. To this end, the number
of comparisons performed by the algorithm on the average to ﬁnd the position of
x is
T(n) =
n
¸
j=1
j
1
n
=
1
n
n
¸
j=1
j =
1
n
n(n + 1)
2
=
n + 1
2
.
This shows that, on the average, the algorithm performs (n + 1)/2 element
comparisons in order to locate x. Hence, the time complexity of Algorithm lin-
earsearch is Θ(n) on the average.
Example 1.31 Consider computing the average number of comparisons per-
formed by Algorithm insertionsort. To simplify the analysis, let us assume
that A[1..n] contains the numbers 1 through n, which implies that all elements
of A are distinct. Furthermore, we will assume that all n! permutations of the
numbers 1, 2, . . . , n are equally likely. Now, consider inserting element A[i] in its
proper position in A[1..i]. If its proper position is j, 1 ≤ j ≤ i, then the number
of comparisons performed in order to insert A[i] in its proper position is i − j if
j = 1 and i − j + 1 if 2 ≤ j ≤ i. Since the probability that its proper position
in A[1..i] is 1/i, the average number of comparisons needed to insert A[i] in its
proper position in A[1..i] is
i −1
i
+
i
¸
j=2
i −j + 1
i
=
i −1
i
+
i−1
¸
j=1
j
i
= 1 −
1
i
+
i −1
2
=
i
2
−
1
i
+
1
2
.
Amortized analysis 47
Thus, the average number of comparisons performed by Algorithm insertion-
sort is
n
¸
i=2

i
2
−
1
i
+
1
2

=
n(n + 1)
4
−
1
2
−
n
¸
i=2
1
i
+
n −1
2
=
n
2
4
+
3n
4
−
n
¸
i=1
1
i
.
Since
ln(n + 1) ≤
n
¸
i=1
1
i
≤ ln n + 1 (Eq. 2.16, page 80),
it follows that the average number of comparisons performed by Algorithm in-
sertionsort is approximately
n
2
4
+
3n
4
−ln n = Θ(n
2
).
Thus, on the average, Algorithm insertionsort performs roughly half the num-
ber of operations performed in the worst case (see Fig. 1.6).
1.13 Amortized analysis
In many algorithms, we may be unable to express the time complexity in
terms of the Θ-notation to obtain an exact bound on the running time.
Therefore, we will be content with the O-notation, which is sometimes
pessimistic. If we use the O-notation to obtain an upper bound on the
running time, the algorithm may be much faster than our estimate even in
the worst case.
Consider an algorithm in which an operation is executed repeatedly with
the property that its running time ﬂuctuates throughout the execution of
the algorithm. If this operation takes a large amount of time occasionally
and runs much faster most of the time, then this is an indication that
amortized analysis should be employed, assuming that an exact bound is
too hard, if not impossible.
In amortized analysis, we average out the time taken by the operation
throughout the execution of the algorithm, and refer to this average as the
amortized running time of that operation. Amortized analysis guarantees
the average cost of the operation, and thus the algorithm, in the worst
case. This is to be contrasted with the average time analysis in which the
average is taken over all instances of the same size. Moreover, unlike the
48 Basic Concepts in Algorithmic Analysis
average case analysis, no assumptions about the probability distribution of
the input are needed.
Amortized time analysis is generally harder than worst case analysis,
but this hardness pays oﬀ when we derive a lower time complexity. A good
example of this analysis will be presented in Sec. 4.3 when we study the
union-ﬁnd algorithms, which is responsible for maintaining a data structure
for disjoint sets. It will be shown that this algorithm runs in time that is
almost linear using amortized time analysis as opposed to a straightforward
bound of O(nlog n). In this section, we present here two simple examples
that convey the essence of amortization.
Example 1.32 Consider the following problem. We have a doubly linked list
(see Sec. 3.2) that initially consists of one node which contains the integer 0. We
have as input an array A[1..n] of n positive integers that are to be processed in
the following way. If the current integer x is odd, then append x to the list. If
it is even, then ﬁrst append x and then remove all odd elements before x in the
list. A sketch of an algorithm for this problem is shown below and is illustrated
in Fig. 1.7 on the input
5 7 3 4 9 8 7 3 .
1. for j ← 1 to n
2. x←A[j]
3. append x to the list
4. if x is even then
5. while pred(x) is odd
6. delete pred(x)
7. end while
8. end if
9. end for
First, 5, 7 and 3 are appended to the list. When 4 is processed, it is inserted
and then 5, 7 and 3 are deleted as shown in Fig. 1.7(f). Next, as shown in
Fig. 1.7(i), after 9 and 8 have been inserted, 9 is deleted. Finally, the elements 7
and 3 are inserted but not deleted, as they do not precede any input integer that
is even.
Now, let us analyze the running time of this algorithm. If the input data
contain no even integers, or if all the even integers are at the beginning, then
no elements are deleted, and hence each iteration of the for loop takes constant
time. On the other hand, if the input consists of n − 1 odd integers followed by
one even integer, then the number of deletions is exactly n − 1, i.e., the number
Amortized analysis 49
(c)
(e)
(g)
(i)
(k)
(b) 0 5 0 5 7
0 5 7 3
4 0 9
4
0 4 8
3 7 8 4 0
(a)
(d)
(f)
(h)
(j)
0
0 7 5 3
4
8
8 7
0
0
0
4
4
9
Fig. 1.7 Illustration of amortized time analysis.
of iterations of the while loop is n − 1. This means that the while loop may
cost Ω(n) time in some iterations. It follows that each iteration of the for loop
takes O(n) time, which results in an overall running time of O(n
2
).
Using amortization, however, we obtain a time complexity of Θ(n) as follows.
The number of insertions is obviously n. As to the number of deletions, we note
that no element is deleted more than once, and thus the number of deletions is
between 0 and n − 1. It follows that the total number of elementary operations
of insertions and deletions altogether is between n and 2n −1. This implies that
the time complexity of the algorithm is indeed Θ(n). It should be emphasized,
however, that in this case we say that the while loop takes constant amortized
time in the worst case. That is, the average time taken by the while loop is
guaranteed to be O(1) regardless of the input.
Example 1.33 Suppose we want to allocate storage for an unknown number
of elements x
1
, x
2
, . . . in a stream of input. One technique to handle the allocation
of memory is to ﬁrst allocate an array A
0
of reasonable size, say m. When this
array becomes full, then upon the arrival of the (m + 1)st element, a new array
A
1
of size 2m is allocated and all the elements stored in A
0
are moved from A
0
to A
1
. Next, the (m+1)st element is stored in A
1
[m+1]. We keep doubling the
size of the newly allocated array whenever it becomes full and a new element is
received, until all elements have been stored.
Suppose, for simplicity, that we start with an array of size 1, that is A
0
consists of only one entry. First, upon arrival of x
1
, we store x
1
in A
0
[1]. When
x
2
is received, we allocate a new array A
1
of size 2, set A
1
[1] to A
0
[1] and store x
2
in A
1
[2]. Upon arrival of the third element x
3
, we allocate a new array A
2
[1..4],
move A
1
[1..2] to A
2
[1..2] and store x
3
in A
2
[3]. The next element, x
4
, will be
stored directly in A
2
[4]. Now since A
2
is full, when x
5
is received, we allocate
a new array A
3
[1..8], move A
2
[1..4] to A
3
[1..4] and store x
5
in A
3
[5]. Next, we
store x
6
, x
7
and x
8
in the remaining free positions of A
3
. We keep doubling the
size of the newly allocated array upon arrival of a new element whenever the
current array becomes full, and move the contents of the current array to the
newly allocated array.
50 Basic Concepts in Algorithmic Analysis
We wish to count the number of element assignments. Suppose, for simplicity,
that the total number of elements received, which is n, is a power of 2. Then the
arrays that have been allocated are A
0
, A
1
, . . . , A
k
, where k = log n. Since x
1
has
been moved k times, x
2
has been moved k −1 times, etc., we may conclude that
each element in ¦x
1
, x
2
, . . . , x
n
¦ has been moved O(k) = O(log n) times. This
implies that the total number of element assignments is O(nlog n).
However, using amortized time analysis, we derive a much tighter bound as
follows. Observe that every entry in each newly allocated array has been assigned
to exactly once. Consequently, the total number of element assignments is equal
to the sum of sizes of all arrays that have been allocated, which is equal to
k
¸
j=0
2
j
= 2
k+1
−1 = 2n −1 = Θ(n) (Eq. 2.9).
Thus, using amortization, it follows that the time needed to store and move each
of the elements x
1
, x
2
, . . . , x
n
is Θ(1) amortized time.
1.14 Input Size and Problem Instance
A measure of the performance of an algorithm is usually a function of its in-
put: its size, order, distribution, etc. The most prominent of these, which is
of interest to us here, is the input size. Using Turing machines as the model
of computation, it is possible, and more convenient indeed, to measure the
input to an algorithm in terms of the number of nonblank cells. This, of
course, is impractical, given that we wish to investigate real world prob-
lems that can be described in terms of numbers, vertices, line segments, and
other varieties of objects. For this reason, the notion of input size belongs
to the practical part of algorithm analysis, and its interpretation has be-
come a matter of convention. When discussing a problem, as opposed to an
algorithm, we usually talk of a problem instance. Thus, a problem instance
translates to input in the context of an algorithm that solves that problem.
For example, we call an array A of n integers an instance of the problem of
sorting numbers. At the same time, in the context of discussing Algorithm
insertionsort, we refer to this array as an input to the algorithm.
The input size, as a quantity, is not a precise measure of the input, and
its interpretation is subject to the problem for which the algorithm is, or
is to be, designed. Some of the commonly used measures of input size are
the following:
Input Size and Problem Instance 51
• In sorting and searching problems, we use the number of entries in
the array or list as the input size.
• In graph algorithms, the input size usually refers to the number of
vertices or edges in the graph, or both.
• In computational geometry, the size of input to an algorithm is
usually expressed in terms of the number of points, vertices, edges,
line segments, polygons, etc.
• In matrix operations, the input size is commonly taken to be the
dimensions of the input matrices.
• In number theory algorithms and cryptography, the number of bits
in the input is usually chosen to denote its length. The number
of words used to represent a single number may also be chosen as
well, as each word consists of a ﬁxed number of bits.
These “heterogeneous” measures have brought about some inconsisten-
cies when comparing the amount of time or space required by two algo-
rithms. For example, an algorithm for adding two n n matrices which
performs n
2
additions sounds quadratic, while it is indeed linear in the
input size.
The the brute-force algorithm for primality testing given in Exam-
ple 1.16. Its time complexity was shown to be O(
√
n). Since this is a
number problem, the time complexity of the algorithm is measured in terms
of the number of bits in the binary representation of n. Since n can be rep-
resented using k = log(n + 1)| bits, the time complexity can be rewritten
as O(
√
n) = O(2
k/2
). Consequently, Algorithm brute-force primali-
tytest is in fact an exponential algorithm.
Now we will compare two algorithms for computing the sum
¸
n
j=1
j. In
the ﬁrst algorithm, which we will call first, the input is an array A[1..n]
with A[j] = j, for each j, 1 ≤ j ≤ n. The input to the second algorithm,
call it second, is just the number n. These two algorithms are shown as
Algorithm first and Algorithm second.
Obviously, both algorithms run in time Θ(n). Clearly, the time com-
plexity of Algorithm first is Θ(n). Algorithm second is designed to solve
a number problem and, as we have stated before, its input size is measured
in terms of the number of bits in the binary representation of the integer
n. Its input consists of k = log (n + 1)| bits. It follows that the time
complexity of Algorithm second is Θ(n) = Θ(2
k
). In other words, it is
considered to be an exponential time algorithm. Ironically, the number of
52 Basic Concepts in Algorithmic Analysis
Algorithm 1.13 first
Input: A positive integer n and an array A[1..n] with A[j] = j, 1 ≤ j ≤ n.
Output:
¸
n
j=1
A[j].
1. sum←0
2. for j ←1 to n
3. sum←sum +A[j]
4. end for
5. return sum
Algorithm 1.14 second
Input: A positive integer n.
Output:
¸
n
j=1
j.
1. sum←0
2. for j ←1 to n
3. sum←sum +j
4. end for
5. return sum
elementary operations performed by both algorithms is the same.
1.15 Exercises
1.1. Let A[1..60] = 11, 12, . . . , 70. How many comparisons are performed by
Algorithm binarysearch when searching for the following values of x?
(a) 33. (b) 7. (c) 70. (d) 77.
1.2. Let A[1..2000] = 1, 2, . . . , 2000. How many comparisons are performed
by Algorithm binarysearch when searching for the following values of
x?
(a) −3. (b) 1. (c) 1000. (d) 4000.
1.3. Draw the decision tree for the binary search algorithm with an input of
(a) 12 elements. (b) 17 elements. (c) 25 elements. (d) 35 elements.
1.4. Illustrate the operation of Algorithm selectionsort on the array
45 33 24 45 12 12 24 12 .
How many comparisons are performed by the algorithm?
1.5. Consider modifying Algorithm selectionsort as shown in Algorithm
Exercises 53
modselectionsort.
Algorithm 1.15 modselectionsort
Input: An array A[1..n] of n elements.
Output: A[1..n] sorted in nondecreasing order.
1. for i ←1 to n −1
2. for j ←i + 1 to n
3. if A[j] < A[i] then interchange A[i] and A[j]
4. end for
5. end for
(a) What is the minimum number of element assignments performed by
Algorithm modselectionsort? When is this minimum achieved?
(b) What is the maximum number of element assignments performed
by Algorithm modselectionsort? Note that each interchange is
implemented using three element assignments. When is this maxi-
mum achieved?
1.6. Illustrate the operation of Algorithm insertionsort on the array
30 12 13 13 44 12 25 13 .
How many comparisons are performed by the algorithm?
1.7. How many comparisons are performed by Algorithm insertionsort
when presented with the input
4 3 12 5 6 7 2 9 ?
1.8. Prove Observation 1.4.
1.9. Which algorithm is more eﬃcient: Algorithm insertionsort or Algo-
rithm selectionsort? What if the input array consists of very large
records? Explain.
1.10. Illustrate the operation of Algorithm bottomupsort on the array
A[1..16] = 11 12 1 5 15 3 4 10 7 2 16 9 8 14 13 6 .
How many comparisons are performed by the algorithm?
1.11. Illustrate the operation of Algorithm bottomupsort on the array
A[1..11] = 2 17 19 5 13 11 4 8 15 12 7 .
How many comparisons are performed by the algorithm?
54 Basic Concepts in Algorithmic Analysis
1.12. Give an array A[1..8] of integers on which Algorithm bottomupsort
performs
(a) the minimum number of element comparisons.
(b) the maximum number of element comparisons.
1.13. Fill in the blanks with either true or false:
f(n) g(n) f = O(g) f = Ω(g) f = Θ(g)
2n
3
+ 3n 100n
2
+ 2n + 100
50n + log n 10n + log log n
50nlog n 10nlog log n
log n log
2
n
n! 5
n
1.14. Express the following functions in terms of the Θ-notation.
(a) 2n + 3 log
100
n.
(b) 7n
3
+ 1000nlog n + 3n.
(c) 3n
1.5
+ (
√
n)
3
log n.
(d) 2
n
+ 100
n
+n!.
1.15. Express the following functions in terms of the Θ-notation.
(a) 18n
3
+ log n
8
.
(b) (n
3
+n)/(n + 5).
(c) log
2
n +
√
n + log log n.
(d) n!/2
n
+n
n/2
.
1.16. Consider the sorting algorithm shown below, which is called bubble-
sort.
Algorithm 1.16 bubblesort
Input: An array A[1..n] of n elements.
Output: A[1..n] sorted in nondecreasing order.
1. i ←1; sorted ←false
2. while i ≤ n −1 and not sorted
3. sorted ←true
4. for j ←n downto i + 1
5. if A[j] < A[j −1] then
6. interchange A[j] and A[j −1]
7. sorted ←false
8. end if
9. end for
10. i ←i + 1
11. end while
Exercises 55
(a) What is the minimum number of element comparisons performed
by the algorithm? When is this minimum achieved?
(b) What is the maximum number of element comparisons performed
by the algorithm? When is this maximum achieved?
(c) What is the minimum number of element assignments performed
by the algorithm? When is this minimum achieved?
(d) What is the maximum number of element assignments performed
by the algorithm? When is this maximum achieved?
(e) Express the running time of Algorithm bubblesort in terms of the
O and Ω notations.
(f) Can the running time of the algorithm be expressed in terms of the
Θ-notation? Explain.
1.17. Find two monotonically increasing functions f(n) and g(n) such that
f(n) ,= O(g(n)) and g(n) ,= O(f(n)).
1.18. Is x = O(xsin x)? Use the deﬁnition of the O-notation to prove your
answer.
1.19. Prove that
¸
n
j=1
j
k
is O(n
k+1
) and Ω(n
k+1
), where k is a positive inte-
ger. Conclude that it is Θ(n
k+1
).
1.20. Let f(n) = ¦1/n + 1/n
2
+ 1/n
3
+ . . .¦. Express f(n) in terms of the
Θ-notation. (Hint: Find a recursive deﬁnition of f(n)).
1.21. Show that n
100
= O(2
n
), but 2
n
,= O(n
100
).
1.22. Show that 2
n
is not Θ(3
n
).
1.23. Is n! = Θ(n
n
)? Prove your answer.
1.24. Is 2
n
2
= Θ(2
n
3
)? Prove your answer.
1.25. Carefully explain the diﬀerence between O(1) and Θ(1).
1.26. Is the function ]log n! O(n), Ω(n), Θ(n)? Prove your answer.
1.27. Can we use the ≺ relation described in Sec. 1.8.6 to compare the order
of growth of n
2
and 100n
2
? Explain.
1.28. Use the ≺ relation to order the following functions by growth rate:
n
1/100
,
√
n, log n
100
, nlog n, 5, log log n, log
2
n, (
√
n)
n
, (1/2)
n
, 2
n
2
, n!.
1.29. Consider the following problem. Given an array A[1..n] of integers, test
each element a in A to see whether it is even or odd. If a is even, then
leave it; otherwise multiply it by 2.
(a) Which one of the O and Θ notations is more appropriate to measure
the number of multiplications? Explain.
(b) Which one of the O and Θ notations is more appropriate to measure
the number of element tests? Explain.
56 Basic Concepts in Algorithmic Analysis
1.30. Give a more eﬃcient algorithm than the one given in Example 1.25.
What is the time complexity of your algorithm?
1.31. Consider Algorithm count4 whose input is a positive integer n.
Algorithm 1.17 count4
1. comment: Exercise 1.31
2. count ←0
3. for i ←1 to ]log n
4. for j ←i to i + 5
5. for k←1 to i
2
6. count ←count + 1
7. end for
8. end for
9. end for
(a) How many times Step 6 is executed?
(b) Which one of the O and Θ notations is more appropriate to express
the time complexity of the algorithm? Explain.
(c) What is the time complexity of the algorithm?
1.32. Consider Algorithm count5 whose input is a positive integer n.
Algorithm 1.18 count5
1. comment: Exercise 1.32
2. count ←0
3. for i ←1 to n
4. j ←]n/2
5. while j ≥ 1
6. count ←count + 1
7. if j is odd then j ←0 else j ←j/2
8. end while
9. end for
(a) What is the maximum number of times Step 6 is executed when n
is a power of 2?
(b) What is the maximum number of times Step 6 is executed when n
is a power of 3?
(c) What is the time complexity of the algorithm expressed in the O-
notation?
(d) What is the time complexity of the algorithm expressed in the Ω-
notation?
Exercises 57
(e) Which one of the O and Θ notations is more appropriate to express
the time complexity of the algorithm?
1.33. Consider Algorithm count6 whose input is a positive integer n.
Algorithm 1.19 count6
1. comment: Exercise 1.33
2. count ←0
3. for i ←1 to n
4. j ←]n/3
5. while j ≥ 1
6. for k←1 to i
7. count ←count + 1
8. end for
9. if j is even then j ←0 else j ←]j/3
10. end while
11. end for
(a) What is the maximum number of times Step 7 is executed when n
is a power of 2?
(b) What is the maximum number of times Step 7 is executed when n
is a power of 3?
(c) What is the time complexity of the algorithm expressed in the O-
notation?
(d) What is the time complexity of the algorithm expressed in the Ω-
notation?
(e) Which one of the O and Θ notations is more appropriate to express
the time complexity of the algorithm?
1.34. Write an algorithm to ﬁnd the maximum and minimum of a sequence of
n integers stored in array A[1..n] such that its time complexity is
(a) O(n).
(b) Ω(nlog n).
1.35. Let A[1..n] be an array of integers, where n > 2. Give an O(1) time
algorithm to ﬁnd an element in A that is neither the maximum nor
the minimum.
1.36. Consider the element uniqueness problem: Given a set of integers, de-
termine whether two of them are equal. Give an eﬃcient algorithm to
solve this problem. Assume that the integers are stored in array A[1..n].
What is the time complexity of your algorithm?
58 Basic Concepts in Algorithmic Analysis
1.37. Give an algorithm that evaluates an input polynomial
a
n
x
n
+a
n−1
x
n−1
+. . . +a
1
x +a
0
for a given value of x in time
(a) Ω(n
2
).
(b) O(n).
1.38. Let S be a set of n positive integers, where n is even. Give an eﬃcient
algorithm to partition S into two subsets S
1
and S
2
of n/2 elements each
with the property that the diﬀerence between the sum of the elements
in S
1
and the sum of the elements in S
2
is maximum. What is the time
complexity of your algorithm?
1.39. Suppose we change the word “maximum” to “minimum” in Exercise 1.38.
Give an algorithm to solve the modiﬁed problem. Compare the time
complexity of your algorithm with that obtained in Exercise 1.38.
1.40. Let m and n be two positive integers. The greatest common divisor of m
and n, denoted by gcd(m, n), is the largest integer that divides both m
and n. For example gcd(12, 18) = 6. Consider Algorithm euclid shown
below, to compute gcd(m, n).
Algorithm 1.20 euclid
Input: Two positive integers m and n.
Output: gcd(m, n).
1. comment: Exercise 1.40
2. repeat
3. r ←n mod m
4. n←m
5. m←r
6. until r = 0
7. return n
(a) Does it matter if in the ﬁrst call gcd(m, n) it happens that n < m?
Explain.
(b) Prove the correctness of Algorithm euclid. (Hint: Make use of
the following theorem: If r divides both m and n, then r divides
m−n).
(c) Show that the running time of Algorithm euclid is maximum if
m and n are two consecutive numbers in the Fibonacci sequence
deﬁned by
f
1
= f
2
= 1; f
n
= f
n−1
+f
n−2
for n > 2.
Bibliographic notes 59
(d) Analyze the running time of Algorithm euclid in terms of n, as-
suming that n ≥ m.
(e) Can the time complexity of Algorithm euclid be expressed using
the Θ-notation? Explain.
1.41. Find the time complexity of Algorithm euclid discussed in Exercise 1.40
measured in terms of the input size. Is it logarithmic, linear, exponen-
tial? Explain.
1.42. Prove that for any constant c > 0, (log n)
c
= o(n).
1.43. Show that any exponential function grows faster than any polynomial
function by proving that for any constants c and d greater than 1,
n
c
= o(d
n
).
1.16 Bibliographic notes
There are several books on the design and analysis of algorithms. These
include, in alphabetical order, Aho, Hopcroft, and Ullman (1974), Baase
(1987), Brassard and Bratley (1988), Brassard and Bratley (1996), Dromey
(1982), Horowitz and Sahni (1978), Hu (1982), Knuth (1968, 1969, 1973),
Manber (1989), Mehlhorn (1984), Moret and Shapiro (1991), Purdom and
Brown (1985), Reingold, Nievergelt, and Deo (1977), Sedgewick (1983) and
Wilf (1986). For a more popular account of algorithms, see Knuth (1977),
Lewis and Papadimitriou (1978) and the two Turing Award Lectures of
Karp (1986) and Tarjan (1987). Some of the more practical aspects of
algorithm design are discussed in Bentley (1982) and Gonnet (1984). Knuth
(1973) discusses in detail the sorting algorithms covered in this chapter.
He gives step-counting analyses. The asymptotic notation was used in
mathematics before the emergence of the ﬁeld of algorithms. Knuth (1976)
gives an account of its history. This article discusses the Ω and Θ notations
and their proper usage and is an attempt to standardize these notations.
Purdom and Brown (1985) presents a comprehensive treatment of advanced
techniques for analyzing algorithms with numerous examples. The main
mathematical aspects of the analysis of algorithms can be found in Greene
and Knuth (1981). Weide (1977) provides a survey of both elementary
and advanced analysis techniques. Hofri (1987) discusses the average-case
analysis of algorithms in detail.
60
Chapter 2
Mathematical Preliminaries
When analyzing an algorithm, the amount of resources required is usually
expressed as a function of the input size. A nontrivial algorithm typi-
cally consists of repeating a set of instructions either iteratively, e.g. by
executing a for or while loop, or recursively by invoking the same algo-
rithm again and again, each time reducing the input size until it becomes
small enough, in which case the algorithm solves the input instance using
a straightforward method. This implies that the amount of resources used
by an algorithm can be expressed in the form of summation or recursive
formula. This mandates the need for the basic mathematical tools that
are necessary to deal with these summations and recursive formulas in the
process of analyzing an algorithm.
In this chapter we review some of the mathematical preliminaries and
discuss brieﬂy some of these mathematical tools that are frequently em-
ployed in the analysis of algorithms.
2.1 Sets, Relations and Functions
When analyzing an algorithm, its input is considered to be a set drawn
from some particular domain, e.g. the set of integers. An algorithm, in
the formal sense, can be thought of as a function, which is a constrained
relation, that maps each possible input to a speciﬁc output. Thus, sets and
functions are at the heart of algorithmic analysis. In this section, we brieﬂy
review some of the basic concepts of sets, relations and functions that arise
naturally in the design and analysis of algorithms. More detailed treatments
can be found in most books on set theory and discrete mathematics.
61
62 Mathematical Preliminaries
2.1.1 Sets
The term set is used to refer to any collection of objects, which are called
members or elements of the set. A set is called ﬁnite if it contains n
elements, for some constant n ≥ 0, and inﬁnite otherwise. Examples of
inﬁnite sets include the set of natural numbers ¦1, 2, . . .¦ and the sets of
integers, rationals and reals.
Informally, an inﬁnite set is called countable if its elements can be listed
as the ﬁrst element, second element, and so on; otherwise it is called un-
countable. For example, the set of integers ¦0, 1, −1, 2, −2, . . .¦ is countable,
while the set of real numbers is uncountable.
A ﬁnite set is described by listing its elements in some way and enclos-
ing this list in braces. If the set is countable, three dots may be used to
indicate that not all the elements have been listed. For example, the set of
integers between 1 and 100 can be stated as ¦1, 2, 3 . . . , 100¦ and the set of
natural numbers can be stated as ¦1, 2, 3 . . .¦. A set may also be denoted
by specifying some property. For example, the set ¦1, 2, . . . , 100¦ can also
be denoted by ¦x [ 1 ≤ x ≤ 100 and x is integer¦. An uncountable set can
only be described this way. For example the set of real numbers between 0
and 1 can be expressed as ¦x [ x is a real number and 0 ≤ x ≤ 1¦. The
empty set is denoted by ¦ ¦ or φ.
If A is a ﬁnite set, then the cardinality of A, denoted by [A[, is the
number of elements in A. We write x ∈ A if x is a member of A, and x / ∈ A
otherwise. We say that a set B is a subset of a set A, denoted by B ⊆ A,
if each element of B is an element of A. If, in addition, B = A, we say
that B is a proper subset of A, and we write B ⊂ A. Thus, ¦a, ¦2, 3¦¦ ⊂
¦a, ¦2, 3¦, b¦, but ¦a, ¦2, 3¦¦ ⊆ ¦a, ¦2¦, ¦3¦, b¦. For any set A, A ⊆ A and
φ ⊆ A. We observe that if A and B are sets such that A ⊆ B and B ⊆ A,
then A = B. Thus, to prove that two sets A and B are equal, we only need
to prove that A ⊆ B and B ⊆ A.
The union of two sets A and B, denoted by A ∪ B, is the set ¦x [ x ∈
A or x ∈ B¦. The intersection of two sets A and B, denoted by A ∩ B, is
the set ¦x [ x ∈ A and x ∈ B¦. The diﬀerence of a set A from a set B,
denoted by A − B, is the set ¦x [ x ∈ A and x / ∈ B¦. The complement of
a set A, denoted by A, is deﬁned as U − A, where U is the universal set
containing A, which is usually understood from the context. If A, B and C
are sets, then A∪(B ∪C) = (A∪B) ∪C, and A∩(B ∩C) = (A∩B) ∩C.
We say that two sets A and B are disjoint if A∩B = φ. The power set of a
Sets, Relations and Functions 63
set A, denoted by P(A), is the set of all subsets of A. Note that φ ∈ P(A)
and A ∈ P(A). If [A[ = n, then [P(A)[ = 2
n
.
2.1.2 Relations
An ordered n-tuple (a
1
, a
2
, . . . , a
n
) is an ordered collection that has a
1
as
its ﬁrst element, a
2
as its second element, . . . , and a
n
as its nth element. In
particular, 2-tuples are called ordered pairs. Let A and B be two sets. The
Cartesian product of A and B, denoted by A B, is the set of all ordered
pairs (a, b) where a ∈ A and b ∈ B. In set notation,
AB = ¦(a, b) [ a ∈ A and b ∈ B¦.
More generally, the Cartesian product of A
1
, A
2
, . . . , A
n
is deﬁned as
A
1
A
2
. . . A
n
= ¦(a
1
, a
2
, . . . , a
n
) [ a
i
∈ A
i
, 1 ≤ i ≤ n¦.
Let A and B be two nonempty sets. A binary relation, or simply a relation,
R from A to B is a set of ordered pairs (a, b) where a ∈ A and b ∈ B, that
is , R ⊆ A B. If A = B, we say that R is a relation on the set A. The
domain of R, sometimes written Dom(R), is the set
Dom(R) = ¦a [ for some b ∈ B (a, b) ∈ R¦.
The range of R, sometimes written Ran(R), is the set
Ran(R) = ¦b [ for some a ∈ A, (a, b) ∈ R¦.
Example 2.1 Let R
1
= ¦(2, 5), (3, 3)¦, R
2
= ¦(x, y) [ x, y are positive integers
and x ≤ y¦ and R
3
= ¦(x, y) [ x, y are real numbers and x
2
+ y
2
≤ 1¦. Then,
Dom(R
1
) = ¦2, 3¦ and Ran(R
1
) = ¦5, 3¦, Dom(R
2
) = Ran(R
2
) is the set of
natural numbers, and Dom(R
3
) = Ran(R
3
) is the set of real numbers in the
interval [−1..1].
Let R be a relation on a set A. R is said to be reﬂexive if (a, a) ∈ R
for all a ∈ A. It is irreﬂexive if (a, a) / ∈ R for all a ∈ A. It is symmetric
if (a, b) ∈ R implies (b, a) ∈ R. It is asymmetric if (a, b) ∈ R implies
(b, a) / ∈ R. It is antisymmetric if (a, b) ∈ R and (b, a) ∈ R implies a = b.
Finally, R is said to be transitive if (a, b) ∈ R and (b, c) ∈ R implies
(a, c) ∈ R. A relation that is reﬂexive, antisymmetric and transitive is
called a partial order.
64 Mathematical Preliminaries
Example 2.2 Let R
1
= ¦(x, y) [ x, y are positive integers and x divides y¦.
Let R
2
= ¦(x, y) [ x, y are integers and x ≤ y¦. Then, both R
1
and R
2
are
reﬂexive, antisymmetric and transitive, and hence both are partial orders.
2.1.2.1 Equivalence relations
A relation R on a set A is called an equivalence relation if it is reﬂexive,
symmetric and transitive. In this case, R partitions A into equivalence
classes C
1
, C
2
, . . . , C
k
such that any two elements in one equivalence class
are related by R. That is, for any C
i
, 1 ≤ i ≤ k, if x ∈ C
i
and y ∈ C
i
,
then (x, y) ∈ R. On the other hand, if x ∈ C
i
and y ∈ C
j
and i = j, then
(x, y) / ∈ R.
Example 2.3 Let x and y be two integers, and let n be a positive integer.
We say that x and y are congruent modulo n, denoted by
x ≡ y (mod n)
if x −y = kn for some integer k. In other words, x ≡ y (mod n) if both x and
y leave the same remainder when divided by n. For example, 13 ≡ 8 (mod 5)
and 13 ≡ −2 (mod 5). Now deﬁne the relation
R = ¦(x, y) [ x, y are integers and x ≡ y (mod n)¦.
Then, R is an equivalence relation. It partitions the set of integers into n classes
C
0
, C
1
, . . . , C
n−1
such that x ∈ C
i
and y ∈ C
i
if and only if x ≡ y (mod n).
2.1.3 Functions
A function f is a (binary) relation such that for every element x ∈ Dom(f)
there is exactly one element y ∈ Ran(f) with (x, y) ∈ f. In this case, one
usually writes f(x) = y instead of (x, y) ∈ f and says that y is the value or
image of f at x.
Example 2.4 The relation ¦(1, 2), (3, 4), (2, 4)¦ is a function, while the re-
lation ¦(1, 2), (1, 4)¦ is not. The relation ¦(x, y) [ x, y are positive integers and
x = y
3
¦ is a function, while the relation ¦(x, y) [ x is a positive integer, y is integer
and x = y
2
¦ is not. In Example 2.1, R
1
is a function, while R
2
and R
3
are not.
Let f be a function such that Dom(f) = A and Ran(f) ⊆ B for some
nonempty sets A and B. We say that f is one to one if for no diﬀerent
Proof Methods 65
elements x and y in A, f(x) = f(y). That is, f is one to one if f(x) = f(y)
implies x = y. We say that f is onto B if Ran(f) = B. f is said to be a
bijection or one to one correspondence between A and B if it is both one to
one and onto B.
2.2 Proof Methods
Proofs constitute an essential component in the design and analysis of al-
gorithms. The correctness of an algorithm and the amount of resources
needed by the algorithm such as its computing time and space usage are all
established by proving postulated assertions. In this section, we brieﬂy re-
view the most common methods of proof used in the analysis of algorithms.
Notation
A proposition or an assertion P is simply a statement that can be either
true or false, but not both. The symbol “” is the negation symbol. For
example, P is the converse of proposition P. The symbols “→”and “↔”
are used extensively in proofs. “→” is read “implies” and “↔” is read “if
and only if”. Thus, if P and Q are two propositions, then the statement
“P → Q” stands for “P implies Q” or “if P then Q”, and the statement
“P ↔ Q”, stands for “P if and only if Q”, that is, “P is true if and
only if Q is true”. The statement“P ↔ Q” is usually broken down into
two implications: “P → Q” and “Q → P”, and each statement is proven
separately. If P → Q, we say that Q is a necessary condition for A, and A
is a suﬃcient condition for B.
2.2.1 Direct proof
To prove that “P → Q”, a direct proof works by assuming that P is true
and then deducing the truth of Q from the truth of P. Many mathematical
proofs are of this type.
Example 2.5 We wish to prove the assertion: If n is an even integer, then n
2
is
an even integer. A direct proof for this claim is as follows. Since n is even, n = 2k
for some integer k. So, n = 4k
2
= 2(2k
2
). It follows that n
2
is an even integer.
66 Mathematical Preliminaries
2.2.2 Indirect proof
The implication “P → Q” is logically equivalent to the contrapositive im-
plication “Q → P”. For example the statement “if it is raining then it
is cloudy” is logically equivalent to the statement “if it is not cloudy then
it is not raining”. Sometimes, proving “if not Q then not P” is much easier
than using a direct proof for the statement “if P then Q”.
Example 2.6 Consider the assertion: If n
2
is an even integer, then n is an
even integer. If we try to use the direct proof technique to prove this theorem,
we may proceed as in the proof in Example 2.5. An alternative approach, which
is much simpler, is to prove the logically equivalent assertion: If n is an odd
integer, then n
2
is an odd integer. We prove the truth of this statement using
the direct proof method as follows. If n is an odd integer, then n = 2k + 1 for
some integer k. Thus, n
2
= (2k +1)
2
= 4k
2
+4k +1 = 2(2k
2
+2k) +1. That is,
n
2
is an odd integer.
2.2.3 Proof by contradiction
This is an extremely powerful method and is widely used to make proofs
short and simple to follow. To prove that the statement “P → Q” is true
using this method, we start by assuming that P is true but Q is false. If
this assumption leads to a contradiction, it means that our assumption that
“Q is false” must be wrong, and hence Q must follow from P. This method
is based on the following logical reasoning. If we know that P → Q is true
and Q is false, then P must be false. So, if we assume at the beginning
that P is true, Q is false, and reach the conclusion that P is false, then we
have that P is both true and false. But P cannot be both true and false,
and hence this is a contradiction. Thus, we conclude that our assumption
that Q is false is wrong, and it must be the case that Q is true after all.
It should be noted that this is not the only contradiction that may result;
for example, after assuming that P is true and Q is false, we may reach
the conclusion that, say, 1 = −1. The following example illustrates this
method of proof. In this example, we make use of the following theorem.
If a, b and c are integers such that a divides both b and c, then a divides
their diﬀerence, that is, a divides b −c.
Example 2.7 We prove the assertion: There are inﬁnitely many primes. We
Proof Methods 67
proceed to prove this assertion by contradiction as follows. Suppose to the con-
trary that there are only k primes p
1
, p
2
, . . . , p
k
, where p
1
= 2, p
2
= 3, p
3
= 5, etc.
and that all other integers greater than 1 are composite. Let n = p
1
p
2
. . . p
k
+ 1
and let p be a prime divisor of n (recall that n is not a prime by assumption since
it is larger than p
k
). Since n is not a prime, one of p
1
, p
2
, . . . , p
k
must divide n.
That is, p is one of the numbers p
1
, p
2
, . . . , p
k
, and hence p divides p
1
p
2
. . . p
k
.
Consequently, p divides n − p
1
p
2
. . . p
k
. But n − p
1
p
2
. . . p
k
= 1, and p does not
divide 1 since it is greater than 1, by deﬁnition of a prime number. This is a
contradiction. It follows that the number of primes is inﬁnite.
The proof of Theorem 2.3 provides an excellent example of the method
of proof by contradiction.
2.2.4 Proof by counterexample
This method provides quick evidence that a postulated statement is false.
It is usually employed to show that a proposition that holds true quite of-
ten is not always true. When we are faced with a problem that requires
proving or disproving a given assertion, we may start by trying to disprove
the assertion with a counterexample. Even if we suspect that the assertion
is true, searching for a counterexample might help to see why a counterex-
ample is impossible. This often leads to a proof that the given statement is
true. In the analysis of algorithms, this method is frequently used to show
that an algorithm does not always produce a result with certain properties.
Example 2.8 Let f(n) = n
2
+ n + 41 be a function deﬁned on the set of
nonnegative integers. Consider the assertion that f(n) is always a prime number.
For example, f(0) = 41, f(1) = 43, . . . f(39) = 1601 are all primes. To falsify this
statement, we only need to ﬁnd one positive integer n such that f(n) is composite.
Since f(40) = 1681 = 41
2
is composite, we conclude that the assertion is false.
Example 2.9 Consider the assertion that

= φ + 1.
Consequently,
f(n) = f(n −1) +f(n −2) ≤ φ
n−2
+φ
n−3
= φ
n−3
(φ + 1) = φ
n−3
φ
2
= φ
n−1
.
Hence, f(n) ≤ φ
n−1
for all n ≥ 1.
Example 2.12 This example shows that if the problem has two or more pa-
rameters, then the choice of the parameter on which to use induction is important.
Let n, m and r denote, respectively, the number of vertices, edges and regions in
an embedding of a planar graph (see Sec. 3.3.2). We will prove Euler’s formula:
n −m+r = 2
stated on page 107. We prove the formula by induction on m, the number of
edges.
Basis step: If m = 1, then there is only one region and two vertices; so 2−1+1 = 2.
Induction step: Suppose the hypothesis holds for 1, 2, . . . , m− 1. We show that
it also holds for m. Let G be a planar graph with n vertices, m − 1 edges
and r regions, and assume that n − (m − 1) + r = 2. Suppose we add one
more edge. Then, we have two cases to consider. If the new edge connects two
vertices that are already in the graph, then one more region will be introduced
and consequently the formula becomes n − m + (r + 1) = n − (m− 1) + r = 2.
If, on the other hand, the added edge connects a vertex in the graph with a new
added vertex, then no more regions are introduced, and the formula becomes
(n + 1) − m + r = n − (m− 1) + r = 2. Thus, the hypothesis holds for m, and
hence for all m ≥ 1.
2.3 Logarithms
Let b be a positive real number greater than 1, x a real number, and suppose
that for some positive real number y we have y = b
x
. Then, x is called the
logarithm of y to the base b, and we write this as
x = log
b
y.
70 Mathematical Preliminaries
Here b is referred to as the base of the logarithm. For any real numbers x
and y greater than 0, we have
log
b
xy = log
b
x + log
b
y,
and
log
b
(c
y
) = y log
b
c, if c > 0.
When b = 2, we will write log x instead of log
2
x.
Another useful base is e, which is deﬁned by
e = lim
n→∞

x
1
1
t
dt.
To convert from one base to another, we use the chain rule:
log
a
x = log
b
x log
a
b or log
b
x =
log
a
x
log
a
b
.
For example,
log x =
ln x
ln 2
and lnx =
log x
log e
.
The following important identity can be proven by taking the logarithms
of both sides:
x
log
b
y
= y
log
b
x
, x, y > 0. (2.2)
2.4 Floor and Ceiling Functions
Let x be a real number. The ﬂoor of x, denoted by x|, is deﬁned as the
greatest integer less than or equal to x. The ceiling of x, denoted by x|,
is deﬁned as the least integer greater than or equal to x. For example,

and log x|| = and log x|.
The following formula follows from Theorem 2.1.
x|/n| = x/n| and x|/n| = x/n|, n a positive integer. (2.3)
For example, if we let x = n/2, then
n/2|/2|/2| = n/4|/2| = n/8|.
2.5 Factorial and Binomial Coeﬃcients
In this section, we brieﬂy list some of the important combinatorial proper-
ties that are frequently used in the analysis of algorithms, especially those
designed for combinatorial problems. We will limit the coverage to per-
mutations and combinations, which lead to the deﬁnitions of factorials and
binomial coeﬃcients.
2.5.1 Factorials
A permutation of n distinct objects is deﬁned to be an arrangement of
the objects in a row. For example there are 6 permutations of a, b and c,
72 Mathematical Preliminaries
namely:
a b c, a c b, b a c, b c a, c a b, c b a.
In general, let there be n > 0 objects, and suppose we want to count the
number of ways to choose k objects and arrange them in a row, where
1 ≤ k ≤ n. Then, there are n choices for the ﬁrst position, n − 1 choices
for the second position,. . . , and n − k + 1 choices for the kth position.
Therefore, the number of ways to choose k ≤ n objects and arrange them
in a row is
n(n −1) . . . (n −k + 1).
This quantity, denoted by P
n
k
, is called the number of permutations of n
objects taken k at a time. When k = n, the quantity becomes
P
n
n
= 1 2 . . . n,
and is commonly called the number of permutations of n objects. Because
of its importance, this quantity is denoted by n!, read “n factorial”. By con-
vention, 0! = 1, which gives the following simple recursive deﬁnition of n!:
0! = 1, n! = n(n −1)! if n ≥ 1.
n! is an extremely fast growing function. For example,
30! = 265252859812191058636308480000000.
A useful approximation to n! is Stirling’s formula:
n! ≈
√
2πn

n
e

n
, (2.4)
where e = 2.7182818 . . . is the base of the natural logarithm. For example,
using Stirling’s formula, we obtain
30! ≈ 264517095922964306151924784891709,
with relative error of about 0.27%.
Factorial and Binomial Coeﬃcients 73
2.5.2 Binomial Coeﬃcients
The possible ways to choose k objects out of n objects, disregarding order,
is customarily called the combinations of n objects taken k at a time. It is
denoted by C
n
k
. For example, the combinations of the letters a, b, c and d
taken three at a time are
a b c, a b d, a c d, b c d.
Since order does not matter here, the combinations of n objects taken k at
a time is equal to the number of permutations of n objects taken k at a
time divided by k!. That is,
C
n
k
=
P
n
k
k!
=
n(n −1) . . . , (n −k + 1)
k!
=
n!
k!(n −k)!
, n ≥ k ≥ 0.
This quantity is denoted by

n
k

, read “n choose k”, which is called the
binomial coeﬃcient. For example, the number of combinations of 4 objects
taken 3 at a time is

4
3

=
4!
3!(4 −3)!
= 4.
Equivalently,

n
k

is the number of k-element subsets in a set of n elements.
For example, the 3-element subsets of the set ¦a, b, c, d¦ are
¦a, b, c¦, ¦a, b, d¦, ¦a, c, d¦, ¦b, c, d¦.
Since the number of ways to choose k elements out of n elements is equal
to the number of ways not to choose n −k elements out of n elements, we
have the following identity.

n
k

=

n
n −k

, in particular

n
n

=

n
0

= 1. (2.5)
The following identity is important:

n
k

=

n −1
k

+

n −1
k −1

. (2.6)
Equation 2.6 can be proven using the following argument. Let A =
¦1, 2, . . . , n¦. Then the k-element subsets can be partitioned into those
74 Mathematical Preliminaries
subsets containing n, and those that do not contain n. The number of sub-
sets not containing n is the number of ways to choose k elements from the
set ¦1, 2, . . . , n−1¦, which is

n −1
k

. The number of subsets containing n
is the number of ways to choose k−1 elements from the set ¦1, 2, . . . , n−1¦,
which is

n −1
k −1

. This establishes the correctness of Eq. 2.6.
The binomial theorem, stated below, is one of the fundamental tools in
the analysis of algorithms. For simplicity, we will state only the special case
when n is a positive integer.
Theorem 2.2 Let n be a positive integer. Then,
(1 +x)
n
=
n
¸
j=0

n
j

x
j
.
If we let x = 1 in Theorem 2.2, we obtain

n
0

+

n
1

+. . . +

n
n

= 2
n
.
In terms of combinations, this identity states that the number of all subsets
of a set of size n is equal to 2
n
, as expected. If we let x = −1 in Theorem 2.2,
we obtain

n
0

−

n
1

+

n
1

−. . . ±

n
n

= 0.
or
n
¸
j even

n
j

=
n
¸
j odd

n
j

.
Letting n = 1, 2, 3, . . . in Theorem 2.2, we obtain the expansions: (1 +
x) = 1 + x, (1 + x)
2
= 1 + 2x + x
2
, (1 + x)
3
= 1 + 3x + 3x
2
+ x
3
, and so
on. If we continue this way indeﬁnitely, we obtain a Pascal triangle, which
is shown in Fig. 2.1. In this triangle, each row after the ﬁrst is computed
from the previous row using Eq. 2.6.
We observe that the binomial coeﬃcients are the digits of the number
(1 +b)
n
written in base b notation, if b is suﬃciently large. For example, in
The pigeonhole principle 75
1
1 1
1 1
1 3 3 1
1
2
4 6
10
1 4
5 1 1 5 10
Fig. 2.1 The ﬁrst six rows of a Pascal triangle.
base 10, we have 11
0
= 1, 11
1
= 11, 11
2
= 121, 11
3
= 1331, 11
4
=
14641.
2.6 The pigeonhole principle
This principle, although easy and intuitive, is extremely powerful and is
indespensible in the analysis of algorithms.
Theorem 2.3 If n balls are distributed into m boxes, then
(1) one box must contain at least n/m| balls, and
(2) one box must contain at most n/m| balls.
Proof. (1) If all boxes have less than n/m| balls, then the total number
of balls is at most
m

n
m
¸
−1

≤ m

n
m
+
m−1
m

−1

= n +m−1 −m < n,
which is a contradiction.
(2) If all boxes have greater than n/m| balls, then the total number of
balls is at least
m

= 2
n+1
−n −2.
2.8.3 Solution of divide-and-conquer recurrences
The main objective of this section is to study some of the techniques speciﬁc
to the solution of the most common divide-and-conquer recurrences that
arise in the analysis of the majority of divide-and-conquer algorithms in
one variable (see Chapter 6). These recurrences take the following form:
f(n) =

d if n ≤ n
0
a
1
f(n/c
1
) +a
2
f(n/c
2
) +. . . +a
p
f(n/c
p
) +g(n) if n > n
0
,
where a
1
, a
2
, . . . , a
p
, c
1
, c
2
, . . . , c
p
and n
0
are nonnegative integers, d a non-
negative constant, p ≥ 1 and g(n) is a function from the set of nonnegative
integers to the set of real numbers. We discuss here three of the most
common techniques of solving divide-and-conquer recurrences.
2.8.3.1 Expanding the recurrence
Perhaps, the most natural approach to solve a recurrence is by expanding it
repeatedly in the obvious way. This method is so mechanical and intuitive
that it virtually does not need any explanation. However, one should keep
in mind that, in some cases, it is time-consuming and, being mechanical,
susceptible to calculation errors. This method is hard to apply on a recur-
rence in which the ratios in the deﬁnition of the function are not equal. An
example of this is given later when we study the substitution method in
Sec. 2.8.3.2.
88 Mathematical Preliminaries
Example 2.23 Consider the recurrence
f(n) =

Θ(n) if a < c.
Θ(nlog n) if a = c.
Θ(n
log
c
a
) if a > c.
Proof. Immediately from Lemma 2.1 and Corollary 2.1. ⁄
2.8.3.2 Substitution
This method is usually employed for proving upper and lower bounds. It
can also be used to prove exact solutions. In this method, we guess a
solution and try to prove it by appealing to mathematical induction.(See
Sec. 2.2.5). Unlike what is commonly done in inductive proofs, here we ﬁrst
proceed to prove the inductive step with one or more unknown constants,
and once the claim is established for f(n), where n is arbitrary, we try
to ﬁne-tune the constant(s), if necessary, in order to make the solution
92 Mathematical Preliminaries
apply to the boundary condition(s) as well. The diﬃculty in this method
is in coming up with an intelligent guess that serves as a tight bound for
the given recurrence. In many instances, however, the given recurrence
resembles another one whose solution is known a priori . This helps in
ﬁnding a starting guess that is reasonably good. The following examples
illustrate this method.
Example 2.24 Consider the recurrence
f(n) =

O(nlog n) if c
1
+c
2
= 1.
Θ(n) if c
1
+c
2
< 1.
Proof. By Example 2.26, f(n) = O(nlog n) if c
1
+c
2
= 1. If c
1
+c
2
< 1,
then by Example 2.27, f(n) = O(n). Since f(n) = Ω(n), it follows that
f(n) = Θ(n). ⁄
2.8.3.3 Change of variables
In some recurrences, it is more convenient if we change the domain of the
function and deﬁne a new recurrence in the new domain whose solution
may be easier to obtain. In the following, we give two examples. The
second example shows that this method is sometimes helpful, as it reduces
the original recurrence to another much easier recurrence.
Example 2.28 Consider the recurrence
f(n) =

d if n = 1
2f(n/2) +bnlog n if n ≥ 2,
which we have solved by expansion in Example 2.23. Here n is a power of 2, so
let k = log n and write n = 2
k
. Then, the recurrence can be rewritten as
f(2
k
) =

d if k = 0
2g(k −1) +b2
k
if k > 0.
This recurrence is of the form of Eq. 2.23 on page 85. Hence, we follow the
procedure outlined in Sec. 2.8.2 to solve this recurrence. If we let
2
k
h(k) = g(k) with h(0) = g(0) = d,
then we have
2
k
h(k) = 2(2
k−1
h(k −1)) +b2
k
.
Dividing both sides by 2
k
yields
h(k) = h(0) +
k
¸
j=1
b = d +bk.
Hence,
g(k) = 2
k
h(k) = d2
k
+bk2
k
.
Substituting n = 2
2
k
, log n = 2
k
and log log n = k yields
f(n) = d log n +b log n log log n.
98 Mathematical Preliminaries
2.9 Exercises
2.1. Let A and B be two sets. Prove the following properties, which are
known as De Morgan’s laws.
(a) A∪ B = A∩ B.
(b) A∩ B = A∪ B.
2.2. Let A, B and C be ﬁnite sets.
(a) Prove the principle of inclusion-exclusion for two sets:
[A∪ B[ = [A[ +[B[ −[A∩ B[.
(b) Prove the principle of inclusion-exclusion for three sets:
[A∪ B ∪ C[ = [A[+[B[+[C[−[A∩ B[−[A∩ C[−[B ∩ C[+[A∩ B ∩ C[.
2.3. Show that if a relation R on a set A is transitive and irreﬂexive, then R
is asymmetric.
2.4. Let R be a relation on a set A. Then, R
2
is deﬁned as ¦(a, b) [ (a, c) ∈
R and (c, b) ∈ R for some c ∈ A¦. Show that if R is symmetric, then R
2
is also symmetric.
2.5. Let R be a nonempty relation on a set A. Show that if R is symmetric
and transitive, then R is not irreﬂexive.
2.6. Let A be a ﬁnite set and P(A) the power set of A. Deﬁne the relation
R on the set P(A) by (X, Y ) ∈ R if and only if X ⊆ Y . Show that R is
a partial order.
2.7. Let A = ¦1, 2, 3, 4, 5¦ and B = A A. Deﬁne the relation R on the set
B by ¦((x, y), (w, z)) ∈ B¦ if and only if xz = yw.
(a) Show that R is an equivalence relation.
(b) Find the equivalence classes induced by R.
2.8. Given the sets A and B and the function f from A to B, determine
whether f is one to one, onto B or both (i.e. a bijection).
(a) A = ¦1, 2, 3, 4, 5¦, B = ¦1, 2, 3, 4¦ and
f =¦(1, 2), (2, 3), (3, 4), (4, 1), (5, 2)¦.
(b) A is the set of integers, B is the set of even integers and f(n) = 2n.
(c) A = B is the set of integers, and f(n) = n
2
.
(d) A = B is the set of real numbers with 0 excluded and f(x) = 1/x.
(e) A = B is the set of real numbers and f(x) = [ x[.
Exercises 99
2.9. A real number r is called rational if r = p/q, for some integers p and
q, otherwise it is called irrational . The numbers 0.25, 1.3333333 . . . are
rational, while π and
√
p, for any prime number p, are irrational. Use
the proof by contradiction method to prove that
√
7 is irrational.
2.10. Prove that for any positive integer n
]log n + 1 = ]log(n + 1).
2.11. Give a counterexample to disprove the assertion given in Example 2.9.
2.12. Use mathematical induction to show that n! > 2
n
for n ≥ 4.
2.13. Use mathematical induction to show that a tree with n vertices has
exactly n −1 edges.
2.14. Prove that φ
n
= φ
n−1
+ φ
n−2
for all n ≥ 2, where φ is the golden ratio
(see Example 2.11).
2.15. Prove that for every positive integer k,
¸
n
i=1
i
k
log i = O(n
k+1
log n).
2.16. Show that
n
¸
j=1
j log j = Θ(n
2
log n)
(a) using algebraic manipulations.
(b) using the method of approximating summations by integration.
2.17. Show that
n
¸
j=1
log(n/j) = O(n),
(a) using algebraic manipulations.
(b) using the method of approximating summations by integration.
2.18. Solve the following recurrence relations.
(a) f(n) = 3f(n −1) for n ≥ 1; f(0) = 5.
(b) f(n) = 2f(n −1) for n ≥ 1; f(0) = 2.
(c) f(n) = 5f(n −1) for n ≥ 1; f(0) = 1.
2.19. Solve the following recurrence relations.
(a) f(n) = 5f(n −1) −6f(n −2) for n ≥ 2; f(0) = 1, f(1) = 0.
(b) f(n) = 4f(n −1) −4f(n −2) for n ≥ 2; f(0) = 6, f(1) = 8.
(c) f(n) = 6f(n −1) −8f(n −2) for n ≥ 2; f(0) = 1, f(1) = 0.
(d) f(n) = −6f(n −1) −9f(n −2) for n ≥ 2; f(0) = 3, f(1) = −3.
(e) 2f(n) = 7f(n −1) −3f(n −2) for n ≥ 2; f(0) = 1, f(1) = 1.
(f) f(n) = f(n −2) for n ≥ 2; f(0) = 5, f(1) = −1.
2.20. Solve the following recurrence relations.
100 Mathematical Preliminaries
(a) f(n) = f(n −1) +n
2
for n ≥ 1; f(0) = 0.
(b) f(n) = 2f(n −1) +n for n ≥ 1; f(0) = 1.
(c) f(n) = 3f(n −1) + 2
n
for n ≥ 1; f(0) = 3.
(d) f(n) = 2f(n −1) +n
2
for n ≥ 1; f(0) = 1.
(e) f(n) = 2f(n −1) +n + 4 for n ≥ 1; f(0) = 4.
(f) f(n) = −2f(n −1) + 2
n
−n
2
for n ≥ 1; f(0) = 1.
(g) f(n) = nf(n −1) + 1 for n ≥ 1; f(0) = 1.
2.21. Consider the following recurrence
f(n) = 4f(n/2) +n for n ≥ 2; f(1) = 1,
where n is assumed to be a power of 2.
(a) Solve the recurrence by expansion.
(b) Solve the recurrence directly by applying Theorem 2.5.
2.22. Consider the following recurrence
f(n) = 5f(n/3) +n for n ≥ 2; f(1) = 1,
where n is assumed to be a power of 3.
(a) Solve the recurrence by expansion.
(b) Solve the recurrence directly by applying Theorem 2.5.
2.23. Consider the following recurrence
f(n) = 9f(n/3) +n
2
for n ≥ 2; f(1) = 1,
where n is assumed to be a power of 3.
(a) Solve the recurrence by expansion.
(b) Solve the recurrence directly by applying Theorem 2.5.
2.24. Consider the following recurrence
f(n) = 2f(n/4) +
√
n for n ≥ 4; f(n) = 1 if n < 4,
where n is assumed to be of the form 2
2
k
, k ≥ 0.
(a) Solve the recurrence by expansion.
(b) Solve the recurrence directly by applying Theorem 2.5.
2.25. Use the substitution method to ﬁnd an upper bound for the recurrence
f(n) = f(]n/2) +f(]3n/4) for n ≥ 4; f(n) = 4 if n < 4.
Express the solution using the O-notation.
Exercises 101
2.26. Use the substitution method to ﬁnd an upper bound for the recurrence
f(n) = f(]n/4) +f(]3n/4) +n for n ≥ 4; f(n) = 4 if n < 4.
Express the solution using the O-notation.
2.27. Use the substitution method to ﬁnd a lower bound for the recurrence in
Exercise 2.25. Express the solution using the Ω-notation.
2.28. Use the substitution method to ﬁnd a lower bound for the recurrence in
Exercise 2.26. Express the solution using the Ω-notation.
2.29. Use the substitution method to solve the recurrence
f(n) = 2f(n/2) +n
2
for n ≥ 2; f(1) = 1,
where n is assumed to be a power of 2. Express the solution using the
Θ-notation.
2.30. Let
f(n) = f(n/2) +n for n ≥ 2; f(1) = 1,
and
g(n) = 2g(n/2) + 1 for n ≥ 2; g(1) = 1,
where n is a power of 2. Is f(n) = g(n)? Prove your answer.
2.31. Use the change of variable method to solve the recurrence
f(n) = f(n/2) +
√
n for n ≥ 4; f(n) = 2 if n < 4,
where n is assumed to be of the form 2
2
k
. Find the asymptotic behavior
of the function f(n)
2.32. Use the change of variable method to solve the recurrence
f(n) = 2f(
√
n) +n for n ≥ 4; f(n) = 1 if n < 4,
where n is assumed to be of the form 2
2
k
. Find the asymptotic behavior
of the function f(n)
2.33. Prove that the solution to the recurrence
f(n) = 2f(n/2) +g(n) for n ≥ 2; f(1) = 1
is f(n) = O(n) whenever g(n) = o(n). For example, f(n) = O(n) if
g(n) = n
1−
, 0 < < 1.
102
Chapter 3
Data Structures
3.1 Introduction
The choice of a suitable data structure can inﬂuence the design of an ef-
ﬁcient algorithm signiﬁcantly. In this chapter, we brieﬂy present some of
the elementary data structures. Our presentation here is not self-contained,
and many details have been omitted. More detailed treatment can be found
in many books on data structures.
3.2 Linked Lists
A linked list consists of a ﬁnite sequence of elements or nodes that contain
information plus (except possibly the last one) a link to another node. If
node x points to node y, then x is called the predecessor of y and y the
successor of x. There is a link to the ﬁrst element called the head of the
list. If there is a link from the last element to the ﬁrst, the list is called
circular. If in a linked list each node (except possibly the ﬁrst one) points
also to its predecessor, then the list is called a doubly linked list. If the
ﬁrst and last nodes are also connected by a pair of links, then we have a
circular doubly linked list. A linked list and its variations are diagrammed
in Fig. 3.1.
The two primary operations on linked lists are insertion and deletion.
Unlike arrays, it costs only a constant amount of time to insert or delete
an element in a linked list. Imposing some restrictions on how a linked list
is accessed results in two fundamental data structures: stacks and queues.
103
104 Data Structures
(a)
(b)
(c)
(d)
Fig. 3.1 Variations of linked lists: (a) Linked list. (b) Circular linked list. (c) Doubly
linked list. (d) Circular doubly linked list.
3.2.1 Stacks and queues
A stack is a linked list in which insertions and deletions are permitted only
at one end, called the top of the stack. It may as well be implemented
as an array. This data structure supports two basic operations: pushing
an element into the stack and popping an element oﬀ the stack. If S is
a stack, the operation pop(S) returns the top of the stack and removes it
permanently. If x is an element of the same type as the elements in S, then
push(S, x) adds x to S and updates the top of the stack so that it points
to x.
A queue is a list in which insertions are permitted only at one end of
the list called its rear, and all deletions are constrained to the other end
called the front of the queue. As in the case of stacks, a queue may also
be implemented as an array. The operations supported by queues are the
same as those for the stack except that a push operation adds an element
at the rear of the queue.
3.3 Graphs
A graph G = (V, E) consists of a set of vertices V = ¦v
1
, v
2
, . . . , v
n
¦ and
a set of edges E. G is either undirected or directed. If G is undirected,
then each edge in E is an unordered pair of vertices. If G is directed,
then each edge in E is an ordered pair of vertices. Figure 3.2 shows an
undirected graph (to the left) and a directed graph (to the right). For ease
Graphs 105
of reference, we will call the undirected and directed graphs in this ﬁgure G
and D, respectively. Let (v
i
, v
j
) be an edge in E. If the graph is undirected,
then v
i
and v
j
are adjacent to each other. If the graph is directed, then v
j
is adjacent to v
i
, but v
i
is not adjacent to v
j
unless (v
j
, v
i
) is an edge in
E. For example, both a and c are adjacent to one another in G, whereas in
D, c is adjacent to a but a is not adjacent to c. The degree of a vertex in
an undirected graph is the number of vertices adjacent to it. The indegree
and outdegree of a vertex v
i
in a directed graph are the number of edges
directed to v
i
and out of v
i
, respectively. For instance, the degree of e in G
is 4, the indegree of c in D is 2 and its outdegree is 1. A path in a graph
from vertex v
1
to vertex v
k
is a sequence of vertices v
1
, v
2
, . . . , v
k
such that
(v
i
, v
i+1
), 1 ≤ i ≤ k −1, is an edge in the graph. The length of a path is the
number of edges in the path. Thus, the length of the path v
1
, v
2
, . . . , v
k
is
k−1. The path is simple if all its vertices are distinct. The path is a cycle if
v
1
= v
k
. An odd-length cycle is one in which the number of edges is odd. An
even-length cycle is deﬁned similarly. For example, a, b, e, a is an odd-length
cycle of length 3 in both G and D. A graph without cycles is called acyclic.
A vertex v is said to be reachable from vertex u if there is a path that
starts at u and ends at v. An undirected graph is connected if every vertex
is reachable from every other vertex, and disconnected otherwise. The
connected components of a graph are the maximal connected subgraphs of
the graph. Thus, if the graph is connected, then it consists of one connected
component, the graph itself. Our example graph G is connected. In the
case of directed graphs, a subgraph is called a strongly connected component
if for every pair of vertices u and v in the subgraph, v is reachable from u
and u is reachable from v. In our directed graph D, there are two strongly
connected components: the subgraph consisting of vertex d only, and the
subgraph consisting of the vertices ¦a, b, c, e¦.
d /
c o
c
d /
c o
c
Fig. 3.2 An undirected and directed graphs.
An undirected graph is said to be complete if there is an edge between
106 Data Structures
each pair of its vertices. A directed graph is said to be complete if there
is an edge from each vertex to all other vertices. Let G = (V, E) be a
complete graph with n vertices. If G is directed, then [E[ = n(n −1). If G
is undirected, then [E[ = n(n −1)/2. The complete undirected graph with
n vertices is denoted by K
n
. An undirected graph G = (V, E) is said to be
bipartite if V can be partitioned into two disjoint subsets X and Y such
that each edge in E has one end in X and the other end in Y . Let m = [X[
and n = [Y [. If there is an edge between each vertex x ∈ X and each vertex
y ∈ Y , then it is called a complete bipartite graph, and is denoted by K
m,n
.
3.3.1 Representation of graphs
A graph G = (V, E) can be conveniently represented by a boolean matrix
M, called the adjacency matrix of G deﬁned as M[i, j] = 1 if and only if
(v
i
, v
j
) is an edge in G. Another representation of a graph is the adjacency
list representation. In this scheme, the vertices adjacent to a vertex are
represented by a linked list. Thus, there are [V [ such lists. Figure 3.3 shows
the adjacency list representations of an undirected and directed graphs.
Clearly, an adjacency matrix of a graph with n vertices has n
2
entries. In
the case of adjacency lists, it costs Θ(m + n) space to represent a graph
with n vertices and m edges.
1
1
2
3
4
5
2 4 5
1 3 5
2 4 5
3 5
1 2 3 4
1
2
3
4
5
4 5
1 3 5
4
3 4
1
5
2 3
4 1
5
2 3
4
Fig. 3.3 An example of adjacency list representation.
Graphs 107
3.3.2 Planar graphs
A graph G = (V, E) is planar if it can be embedded in the plane with-
out edge crossings. Figure 3.4(a) shows an example of a planar graph.
This graph is planar because it can be embedded in the plane as shown in
Fig. 3.4(b).
(a) (b)
Fig. 3.4 An example of a planar graph.
The importance of planar graphs comes from the relationship between
their number of vertices, number of edges and number of regions. Let n, m
and r denote, respectively, the number of vertices, edges and regions in any
embedding of a planar graph. Then, these three parameters are related by
Euler’s formula
n −m+r = 2
or
m = n +r −2.
Moreover, we have
m ≤ 3n −6.
The equality is attained if the graph is triangulated, i.e., each one of its
regions (including the unbounded region) is triangular. The graph shown
in Fig. 3.4(b) is triangulated and hence the relation m = 3n − 6 holds
for this graph. The above relationships imply that in any planar graph
m = O(n). Thus, the amount of space needed to store a planar graph is
only Θ(n). This is to be contrasted with complete graphs, which require
an amount of space in the order of Θ(n
2
).
108 Data Structures
3.4 Trees
A free tree (or simply a tree) is a connected undirected graph that contains
no cycles. A forest is a vertex-disjoint collection of trees, i.e., they do not
have vertices in common.
Theorem 3.1 If T is a tree with n vertices, then
(a) Any two vertices of T are connected by a unique path.
(b) T has exactly n −1 edges.
(c) The addition of one more edge to T creates a cycle.
Since the number of edges in a tree is n−1, when analyzing the time or
space complexity in the context of trees, the number of edges is insigniﬁcant.
3.5 Rooted Trees
A rooted tree T is a (free) tree with a distinguished vertex r called the root
of T. This imposes an implicit direction on the path from the root to every
other vertex. A vertex v
i
is the parent of vertex v
j
in T if v
i
is on the
path from the root to v
j
and is adjacent to v
j
. In this case, v
j
is a child
of v
i
. The children of a vertex are called siblings. A leaf of a rooted tree
is a vertex with no children; all other vertices are called internal vertices.
A vertex u on the path from the root to a vertex v is an ancestor of v.
If u = v, then u is a proper ancestor of v. A vertex w on the path from
a vertex v to a leaf is a descendant of v. If w = v, then w is a proper
descendant of v. The subtree rooted at a vertex v is the tree consisting of
v and its proper descendants. The depth of a vertex v in a rooted tree is
the length of the path from the root to v. Thus, the depth of the root is 0.
The height of a vertex v is deﬁned as the length of the longest path from v
to a leaf. The height of a tree is the height of its root.
Example 3.1 Consider the rooted tree T shown in Fig. 3.5. Its root is the
vertex labeled a. b is the parent of e and f, which in turn are the children of b.
b, c and d are siblings. e, f, g and d are leaves; the others are internal vertices. e
is a (proper) descendant of both a and b, which in turn are (proper) ancestors of
e. The subtree rooted at b is the tree consisting of b and its children. The depth
of g is 2; its height is 0. Since the distance from a to g is 2, and no other path
Binary Trees 109
o
c d /
c ;
o
Fig. 3.5 An example of a rooted tree.
from a to a leaf is longer than 2, the height of a is 2. It follows that the height of
T is the height of its root, which is 2.
3.5.1 Tree traversals
There are several ways in which the vertices of a rooted tree can be sys-
tematically traversed or ordered. The three most important orderings are
preorder, inorder and postorder. Let T be a tree with root r and subtrees
T
1
, T
2
, . . . , T
n
.
• In a preorder traversal of the vertices of T, we visit the root r
followed by visiting the vertices of T
1
in preorder, then the vertices
of T
2
in preorder, and so on up to the vertices of T
n
in preorder.
• In an inorder traversal of the vertices of T, we visit the vertices
of T
1
in inorder, then the root r, followed by the vertices of T
2
in
inorder, and so on up to the vertices of T
n
in inorder.
• In a postorder traversal of the vertices of T, we visit the vertices of
T
1
in postorder, then the vertices of T
2
in postorder, and so on up
to the vertices of T
n
in postorder, and ﬁnally we visit r.
3.6 Binary Trees
A binary tree is a ﬁnite set of vertices that is either empty or consists of a
root r and two disjoint binary trees called the left and right subtrees. The
roots of these subtrees are called the left and right children of r. Binary
trees diﬀer from rooted trees in two important ways. First, a binary tree
110 Data Structures
may be empty while a rooted tree cannot be empty. Second, the distinction
of left and right subtrees causes the two binary trees shown in Fig. 3.6(a)
and (b) to be diﬀerent, and yet as rooted trees, they are indistinguishable.
&
·
&
·
(a) (b)
Fig. 3.6 Two diﬀerent binary trees.
All other deﬁnitions of rooted trees carry over to binary trees. A binary
tree is said to be full if each internal vertex has exactly two children. A
binary tree is called complete if it is full and all its leaves have the same
depth, i.e., are on the same level. Figure 3.7 shows a complete binary tree.
The set of vertices in a binary tree is partitioned into levels, with each level
consisting of those vertices with the same depth (see Fig. 3.7).
level 0
level 1
level 2
level 3
Fig. 3.7 A complete binary tree.
Thus, level i consists of those vertices of depth i. We deﬁne a binary
tree to be almost-complete if it is complete except that possibly one or
more leaves that occupy the rightmost positions may be missing. Hence,
by deﬁnition, an almost-complete binary tree may be complete. Figure 3.8
shows an almost-complete binary tree. In this tree the three rightmost
leaves are missing.
A complete (or almost-complete) binary tree with n vertices can be
represented eﬃciently by an array A[1..n] that lists its vertices according
to the following simple relationship: The left and right children (if any) of
Binary Trees 111
Fig. 3.8 An almost-complete binary tree.
a vertex stored in A[j] are stored in A[2j] and A[2j + 1], respectively, and
the parent of a vertex stored in A[j] is stored in A[j/2|].
3.6.1 Some quantitative aspects of binary trees
In the following observations, we list some useful relationships between the
levels, number of vertices and height of a binary tree.
Observation 3.1 In a binary tree, the number of vertices at level j is at
most 2
j
.
Observation 3.2 Let n be the number of vertices in a binary tree T of
height k. Then,
n ≤
k
¸
j=0
2
j
= 2
k+1
−1.
The equality holds if T is complete. If T is almost-complete, then we have
2
k
≤ n ≤ 2
k+1
−1.
Observation 3.3 The height of any binary tree with n vertices is at least
log n| and at most n −1.
Observation 3.4 The height of a complete or almost-complete binary
tree with n vertices is log n|.
Observation 3.5 In a full binary tree, the number of leaves is equal to
the number of internal vertices plus one.
112 Data Structures
3.6.2 Binary search trees
A binary search tree is a binary tree in which the vertices are labeled with
elements from a linearly ordered set in such a way that all elements stored
in the left subtree of a vertex v are less than the element stored at vertex v,
and all elements stored in the right subtree of a vertex v are greater than
the element stored at vertex v. This condition, which is called the binary
search tree property, holds for every vertex of a binary search tree. The
representation of a set by a binary search tree is not unique; in the worst
case it may be a degenerate tree, i.e., a tree in which each internal vertex
has exactly one child. Figure 3.9 shows two binary search trees representing
the same set.
4
3
1 6
8
9
9
8
6
3
4 1
Fig. 3.9 Two binary search trees representing the same set.
The operations supported by this data structure are insertion, deletion,
testing for membership and retrieving the minimum or maximum.
3.7 Exercises
3.1. Write an algorithm to delete an element x, if it exists, from a doubly-
linked list L. Assume that the variable head points to the ﬁrst element
in the list and the functions pred(y) and next(y) return the predecessor
and successor of node y, respectively.
3.2. Give an algorithm to test whether a list has a repeated element.
3.3. Rewrite Algorithm insertionsort so that its input is a doubly linked
list of n elements instead of an array. Will the time complexity change?
Is the new algorithm more eﬃcient?
3.4. A polynomial of the form p(x) = a
1
x
b
1
+ a
2
x
b
2
+ . . . + a
n
x
b
n
, where
b
1
> b
2
> . . . > b
n
≥ 0, can be represented by a linked list in which each
record has three ﬁelds for a
i
, b
i
and the link to the next record. Give
an algorithm to add two polynomials using this representation. What is
the running time of your algorithm?
Exercises 113
3.5. Give the adjacency matrix and adjacency list representations of the graph
shown in Fig. 3.5.
3.6. Describe an algorithm to insert and delete edges in the adjacency list
representation for
(a) a directed graph.
(b) an undirected graph.
3.7. Let S
1
be a stack containing n elements. Give an algorithm to sort the
elements in S
1
so that the smallest element is on top of the stack after
sorting. Assume you are allowed to use another stack S
2
as a temporary
storage. What is the time complexity of your algorithm?
3.8. What if you are allowed to use two stacks S
2
and S
3
as a temporary
storage in Exercise 3.7?
3.9. Let G be a directed graph with n vertices and m edges. When is it the
case that the adjacency matrix representation is more eﬃcient that the
adjacency lists representation? Explain.
3.10. Prove that a graph is bipartite if and only if it has no odd-length cycles.
3.11. Draw the almost-complete binary tree with
(a) 10 nodes.
(b) 19 nodes.
3.12. Prove Observation 3.1.
3.13. Prove Observation 3.2.
3.14. Prove Observation 3.4.
3.15. Prove Observation 3.3.
3.16. Prove Observation 3.5.
3.17. Is a tree a bipartite graph? Prove your answer (see Exercise 3.10).
3.18. Let T be a nonempty binary search tree. Give an algorithm to
(a) return the minimum element stored in T.
(b) return the maximum element stored in T.
3.19. Let T be a nonempty binary search tree. Give an algorithm to list all
the elements in T in increasing order. What is the time complexity of
your algorithm?
3.20. Let T be a nonempty binary search tree. Give an algorithm to delete
an element x from T, if it exists. What is the time complexity of your
algorithm?
3.21. Let T be binary search tree. Give an algorithm to insert an element x in
its proper position in T. What is the time complexity of your algorithm?
114 Data Structures
3.22. What is the time complexity of deletion and insertion in a binary search
tree? Explain.
3.23. When discussing the time complexity of an operation in a binary search
tree, which of the O and Θ notations is more appropriate? Explain.
3.8 Bibliographic notes
This chapter outlines some of the basic data structures that are frequently
used in the design and analysis of algorithms. More detailed treatment can
be found in many books on data structures. These include, among others,
Aho, Hopcroft and Ullman (1983), Gonnet (1984), Knuth (1968), Knuth
(1973), Reingold and Hansen (1983), Standish (1980), Tarjan (1983) and
Wirth (1986). The deﬁnitions in this chapter conform with those in Tarjan
(1983). The adjacency lists data structure was suggested by Tarjan and is
described in Tarjan (1972) and Hopcroft and Tarjan (1973).
Chapter 4
Heaps and the Disjoint Sets Data
Structures
4.1 Introduction
In this chapter, we investigate two major data structures that are more
sophisticated than those in Chapter 3. They are fundamental in the design
of eﬃcient algorithms. Moreover, these data structures are interesting in
their own right.
4.2 Heaps
In many algorithms, there is the need for a data structure that supports
the two operations: Insert an element and ﬁnd the element of maximum
value. A data structure that supports both of these operations is called a
priority queue. If a regular queue is used, then ﬁnding the largest element
is expensive, as this requires searching the entire queue. If a sorted array is
used, then insertion is expensive, as it may require shifting a large portion
of the elements. An eﬃcient implementation of a priority queue is to use a
simple data structure called a heap.
Deﬁnition 4.1 A (binary) heap is an almost-complete binary tree (see
Sec. 3.6) with each node satisfying the heap property: If v and p(v) are a
node and its parent, respectively, then the key of the item stored in p(v) is
not less than the key of the item stored in v.
A heap data structure supports the following operations:
115
116 Heaps and the Disjoint Sets Data Structures
• delete-max[H]: Delete and return an item of maximum key from a
nonempty heap H.
• insert[H, x]: Insert item x into heap H.
• delete[H, i]: Delete the ith item from heap H.
• makeheap[A]: Transform array A into a heap.
Thus, the heap property implies that the keys of the elements along
every path from the root to a leaf are arranged in nonincreasing order. As
described in Sec. 3.6, a heap T (being an almost-complete binary tree) with
n nodes can be represented by an array H[1..n] in the following way:
• The root of T is stored in H[1].
• Suppose that a node x in T is stored in H[j]. If it has a left child,
then this child is stored in H[2j]. If it (also) has a right child, then
this child is stored in H[2j + 1].
• The parent of element H[j] that is not the root of the tree is stored
in H[j/2|].
Note that if a node in a heap has a right child, then it must also have
a left child. This follows from the deﬁnition of an almost-complete binary
tree. Consequently, a heap can be viewed as a binary tree, while it is
in fact an array H[1..n] with the property that for any index j, 2 ≤ j ≤
n, key(H[j/2|]) ≥ key(H[j]). Figure 4.1 shows an example of a heap
in both tree and array representations. To simplify this and subsequent
ﬁgures, we will treat the keys of the items stored in a heap as if they
themselves are the items. In Fig. 4.1, we note that if the tree nodes are
numbered from 1 to n in a top-down and left to right manner, then each
entry H[i] is represented in the corresponding tree by the node numbered
i. This numbering is indicated in the ﬁgure by the labels next to the tree
nodes. Thus, using this method, given a heap as an array, we can easily
construct its corresponding tree and vice-versa.
4.2.1 Operations on heaps
Before describing the main heap operations, we ﬁrst present two secondary
operations that are used as subroutines in the algorithms that implement
heap operations.
Heaps 117
10
20
17
9
10
11
4
5
3
7
5
1
2
3
4
5
6
7
8
9
3 7 5
10 11 4 5
9 17
20
1
2
3
4
5
6
7
8
9 10
Fig. 4.1 A heap and its array representation.
Sift-up
Suppose that for some i > 1, H[i] is changed to an element whose key is
greater than the key of its parent. This violates the heap property and,
consequently, the data structure is no longer a heap. To restore the heap
property, an operation called sift-up is needed to move the new item up to
its proper position in the binary tree so that the heap property is restored.
The sift-up operation moves H[i] up along the unique path from H[i] to the
root until its proper location along this path is found. At each step along
the path, the key of H[i] is compared with the key of its parent H[i/2|].
This is described more precisely in Procedure sift-up.
Procedure sift-up
Input: An array H[1..n] and an index i between 1 and n.
Output: H[i] is moved up, if necessary, so that it is not larger than its parent.
1. done←false
2. if i = 1 then exit ¦node i is the root¦
3. repeat
4. if key(H[i]) > key(H[]i/2]) then interchange H[i] and H[]i/2]
5. else done←true
6. i ←]i/2
7. until i = 1 or done
Example 4.1 Suppose the key stored in the 10th position of the heap shown
118 Heaps and the Disjoint Sets Data Structures
in Fig. 4.1 is changed from 5 to 25. This will violate the heap property, as the
new key 25 is now greater than the key stored in its parent node, namely 11. To
restore the heap property, we apply the sift-up operation to the tree starting from
that node in which 25 is stored. This action is depicted in Fig. 4.2. As shown in
the ﬁgure, 25 is moved up to the root.
3 7
25
10
11
4
5
9
17
20
Fig. 4.2 An example of the sift-up operation.
Sift-down
Suppose that i ≤ n/2| and the key of the element stored in H[i] is changed
to a value that is less than the maximum of the keys at H[2i] and H[2i+1] (if
H[2i+1] exists). This violates the heap property and the tree is no longer a
representation of a heap. To restore the heap property, an operation called
sift-down is needed to “percolate” H[i] down the binary tree until its proper
location is found. At each step along the path, its key is compared with
the maximum of the two keys stored in its children nodes (if any). This is
described more formally in Procedure sift-down.
Example 4.2 Suppose we change the key 17 stored in the second position
of the heap shown in Fig. 4.1 to 3. This will violate the heap property, as the
new key 3 is now less than the maximum of the two keys stored in its children
nodes, namely 11. To restore the heap property, we apply the sift-down operation
starting from that node in which 3 is stored. This action is depicted in Fig. 4.3.
As shown in the ﬁgure, 3 is percolated down until its proper position is found.
Now, using these two procedures, it is fairly easy to write the algorithms
for the main heap operations.
Heaps 119
Procedure sift-down
Input: An array H[1..n] and an index i between 1 and n.
Output: H[i] is percolated down, if necessary, so that it is not smaller
than its children.
1. done ← false
2. if 2i > n then exit ¦node i is a leaf¦
3. repeat
4. i ←2i
5. if i + 1 ≤ n and key(H[i + 1]) > key(H[i]) then i ←i + 1
6. if key(H[]i/2]) < key(H[i]) then interchange H[i] and H[]i/2]
7. else done ←true
8. end if
9. until 2i > n or done
3 7
5 10
11
4 5
9
3
20
Fig. 4.3 An example of the sift-down operation.
Insert
To insert an element x into a heap H, append x to the end of H after
its size has been increased by 1, and then sift x up, if necessary, until
the heap property is satisﬁed. This is described in Algorithm insert. By
Observation 3.4, if n is the size of the new heap, then the height of the heap
tree is log n|. It follows that the time required to insert one element into
a heap of size n is O(log n).
Delete
To delete an element H[i] from a heap H of size n, replace H[i] by H[n],
decrease the heap size by 1, and then sift H[i] up or down, if necessary,
depending on the value of its key relative to the keys stored in its parent
and children nodes, until the heap property is satisﬁed. This is described in
120 Heaps and the Disjoint Sets Data Structures
Algorithm 4.1 insert
Input: A heap H[1..n] and a heap element x.
Output: A new heap H[1..n + 1] with x being one of its elements.
1. n←n + 1 ¦increase the size of H¦
2. H[n] ←x
3. sift-up(H, n)
Algorithm delete. Since, by Observation 3.4, the height of the heap tree
is log n|, it follows that the time required to delete a node from a heap of
size n is O(log n).
Algorithm 4.2 delete
Input: A nonempty heap H[1..n] and an index i between 1 and n.
Output: A new heap H[1..n −1] after H[i] is removed.
1. x←H[i]; y←H[n]
2. n←n −1 ¦decrease the size of H¦
3. if i = n + 1 then exit ¦done¦
4. H[i] ←y
5. if key(y) ≥ key(x) then sift-up(H, i)
6. else sift-down(H, i)
7. end if
Delete-max
This operation deletes and returns an item of maximum key in a nonempty
heap H. It costs Θ(1) time to return the element with maximum key
in a heap, as it is the root of the tree. However, since deleting the root
destroys the heap, more work is needed to restore the heap data structure.
A straightforward implementation of this operation makes use of the delete
operation: Simply return the element stored in the root and delete it from
the heap. The method for this operation is given in Algorithm deletemax.
Obviously, its time complexity is that of the delete operation, i.e., O(log n).
4.2.2 Creating a heap
Given an array A[1..n] of n elements, it is easy to construct a heap out of
these elements by starting from an empty heap and successively inserting
Heaps 121
Algorithm 4.3 deletemax
Input: A heap H[1..n].
Output: An element x of maximum key is returned and deleted from the heap.
1. x←H[1]
2. delete(H, 1)
3. return x
each element until A is transformed into a heap. Since inserting the jth key
costs O(log j), the time complexity of creating a heap using this method is
O(nlog n) (see Example 1.12).
Interestingly, it turns out that a heap can be created from n elements in
Θ(n) time. In what follows, we give the details of this method. Recall that
the nodes of the tree corresponding to a heap H[1..n] can conveniently be
numbered from 1 to n in a top-down left to right manner. Given this num-
bering, we can transform an almost-complete binary tree with n nodes into
a heap H[1..n] as follows. Starting from the last node (the one numbered
n) to the root (node number 1), we scan all these nodes one by one, each
time transforming, if necessary, the subtree rooted at the current node into
a heap.
Example 4.3 Figure 4.4 provides an example of the linear time algorithm for
transforming an array A[1..n] into a heap. An input array and its tree represen-
tation are shown in Fig. 4.4(a). Each subtree consisting of only one leaf is already
a heap, and hence the leaves are skipped. Next, as shown in Fig. 4.4(b), the two
subtrees rooted at the 4th and 5th nodes are not heaps, and hence their roots
are sifted down in order to transform them into heaps. At this point, all subtrees
rooted at the second and third levels are heaps. Continuing this way, we adjust
the two subtrees rooted at the third and second nodes in the ﬁrst level, so that
they conform to the heap property. This is shown in Figs. 4.4(c)-(d). Finally,
we move up to the topmost level and percolate the element stored in the root
node down to its proper position. The resulting tree, which is now a heap, and
its array representation are shown in Fig. 4.4(e).
To this end, we have shown how to work with trees. Performing the
same procedure directly on the input array is fairly easy. Let A[1..n] be
the given array and T the almost-complete binary tree corresponding to A.
122 Heaps and the Disjoint Sets Data Structures
(a)
10
4
3
8
10
11
13
7
30
17
26
1
2
3
4
5
6
7
8
9
1
3
8
2
4
5
3
11 10
10 9
17
8
30
7
7
6
13
26
4
(d)
30
11
26
3 10
17
8
9
10
4
2
5
3
11
26
17 10
30
8
9
10
4 5
17
10
8
26
13
7
30
3
11
(e)
8
1
3 13
2
4
5
26
11 17
10
4
9
3 10
7
7
6
8
30
10 4
1
2
3
4
5
6
7
8
9
2
(b)
17 10
30
9 8
4
17 30
10
9 8
4
11
26
5
10
26
11
5
10
(c)
8
13
7
6
7
3
13
8
7
6
7
3
Fig. 4.4 An example of creating a heap.
First, we note that the elements
A[n/2| + 1], A[n/2| + 2], . . . , A[n]
correspond to the leaves of T, and therefore we may start adjusting the
Heaps 123
array at A[n/2|] and continue the adjustments at
A[n/2| −1], A[n/2| −2], . . . , A[1].
Once the subtree rooted at A[1], which is T itself, is adjusted, the resulting
array is the desired heap. Algorithm makeheap constructs a heap whose
items are the elements stored in an array A[1..n].
Algorithm 4.4 makeheap
Input: An array A[1..n] of n elements.
Output: A[1..n] is transformed into a heap
1. for i ←]n/2 downto 1
2. sift-down(A, i)
3. end for
The running time of Algorithm makeheap is computed as follows. Let
T be the almost-complete binary tree corresponding to the array A[1..n].
Then, by Observation 3.4, the height of T is k = log n|. Let A[j] cor-
responds to the jth node in level i of the tree. The number of iterations
executed by Procedure sift-down when invoked by the statement sift-
down(A, j) is at most k − i. Since there are exactly 2
i
nodes on level
i, 0 ≤ i < k, the total number of iterations is bounded above by
k−1
¸
i=0
(k −i)2
i
= 2
0
(k) + 2
1
(k −1) + 2
2
(k −2) +. . . + 2
k−1
(1)
= 1(2
k−1
) + 2(2
k−2
) +. . . +k(2
k−k
)
=
k
¸
i=1
i2
(k−i)
= 2
k
k
¸
i=1
i/2
i
≤ n
k
¸
i=1
i/2
i
< 2n.
The last equality follows from Eq. 2.14 on page 78. Since there are at
124 Heaps and the Disjoint Sets Data Structures
most two element comparisons in each iteration of Procedure sift-down,
the total number of element comparisons is bounded above by 4n. More-
over, since there is at least one iteration in each call to Procedure sift-
down, the minimum number of element comparisons is 2n/2| ≥ n − 1.
Thus, we have the following theorem:
Theorem 4.1 Let C(n) be the number of element comparisons performed
by Algorithm makeheap for the construction of a heap of n elements.
Then, n −1 ≤ C(n) < 4n. Hence, the algorithm takes Θ(n) time and Θ(1)
space to construct a heap out of n elements.
4.2.3 Heapsort
We now turn our attention to the problem of sorting by making use of the
heap data structure. Recall that Algorithm selectionsort sorts an ar-
ray of n elements by ﬁnding the minimum in each of the n − 1 iterations.
Thus, in each iteration, the algorithm searches for the minimum among
the remaining elements using linear search. Since searching for the mini-
mum using linear search costs Θ(n) time, the algorithm takes Θ(n
2
) time.
It turns out that by choosing the right data structure, Algorithm selec-
tionsort can be improved substantially. Since we have at our disposal
the heap data structure with the delete-max operation, we can exploit it
to obtain an eﬃcient algorithm. Given an array A[1..n], we sort its ele-
ments in nondecreasing order eﬃciently as follows. First, we transform A
into a heap with the property that the key of each element is the element
itself, i.e., key(A[i]) = A[i], 1 ≤ i ≤ n. Next, since the maximum of the
entries in A is now stored in A[1], we may interchange A[1] and A[n] so that
A[n] is the maximum element in the array. Now, the element stored in A[1]
may be smaller than the element stored in one of its children. Therefore,
we use Procedure sift-down to transform A[1..n − 1] into a heap. Next,
we interchange A[1] with A[n − 1] and adjust the array A[1..n − 2] into a
heap. This process of exchanging elements and adjusting heaps is repeated
until the heap size becomes 1, at which point A[1] is minimum. The formal
description is shown in Algorithm heapsort.
An important advantage of this algorithm is that it sorts in place, i.e., it
needs no auxiliary storage. In other words, the space complexity of Algo-
rithm heapsort is Θ(1). The running time of the algorithm is computed as
Disjoint Sets Data Structures 125
Algorithm 4.5 heapsort
Input: An array A[1..n] of n elements.
Output: Array A sorted in nondecreasing order
1. makeheap(A)
2. for j ←n downto 2
3. interchange A[1] and A[j]
4. sift-down(A[1..j −1], 1)
5. end for
follows. By Theorem 4.1, creating the heap costs Θ(n) time. The sift-down
operation costs O(log n) time and is repeated n − 1 times. It follows that
the time required by the algorithm to sort n elements is O(nlog n). This
implies the following theorem:
Theorem 4.2 Algorithm heapsort sorts n elements in O(nlog n) time
and Θ(1) space.
4.2.4 Min and Max Heaps
So far we have viewed the heap as a data structure whose primary operation
is retrieving the element with maximum key. The heap can trivially be
modiﬁed so that the element with minimum key value is stored in the
root instead. In this case, the heap property mandates that the key of
the element stored in a node other than the root is greater than or equal
to the key of the element stored in its parent. These two types of heaps
are commonly referred to as max-heaps and min-heaps. The latter is not
less important than the former, and they are both used quite often in
optimization algorithms. It is customary to refer to either one of them as
a “heap” and which one is meant is understood from the context in which
it is used.
4.3 Disjoint Sets Data Structures
Suppose we are given a set S of n distinct elements. The elements are
partitioned into disjoint sets. Initially, each element is assumed to be in a set
by itself. A sequence σ of m union and ﬁnd operations, which will be deﬁned
126 Heaps and the Disjoint Sets Data Structures
below, is to be executed so that after each union instruction, two disjoint
subsets are combined into one subset. Observe that the number of unions
is at most n − 1. In each subset, a distinguished element will serve as the
name of the set or set representative. For example, if S = ¦1, 2, . . . , 11¦ and
there are 4 subsets ¦1, 7, 10, 11¦, ¦2, 3, 5, 6¦, ¦4, 8¦ and ¦9¦, these subsets
may be labeled as 1, 3, 8 and 9, in this order. The ﬁnd operation returns
the name of the set containing a particular element. For example, executing
the operation ﬁnd(11) returns 1, the name of the set containing 11. These
two operations are deﬁned more precisely as follows.
• find(x) : Find and return the name of the set containing x.
• union(x, y) : Replace the two sets containing x and y by their
union. The name of the union set is either the name of the old
set containing x or the name of the old set containing y; it will be
determined later.
The goal is to design eﬃcient algorithms for these two operations. To
achieve this, we need a data structure that is both simple and at the same
time allows for the eﬃcient implementation of the union and ﬁnd opera-
tions. A data structure that is both simple and leads to eﬃcient implemen-
tation of these two operations is to represent each set as a rooted tree with
data elements stored in its nodes. Each element x other than the root has
a pointer to its parent p(x) in the tree. The root has a null pointer, and it
serves as the name or set representative of the set. This results in a forest
in which each tree corresponds to one set.
For any element x, let root(x) denote the root of the tree containing x.
Thus, find(x) always returns root(x). As the union operation must have
as its arguments the roots of two trees, we will assume that for any two
elements x and y, union(x, y) actually means union(root(x), root(y)).
If we assume that the elements are the integers 1, 2, . . . , n, the forest
can conveniently be represented by an array A[1..n] such that A[j] is the
parent of element j, 1 ≤ j ≤ n. The null parent can be represented by the
number 0. Figure 4.5(a) shows four trees corresponding to the four sets
¦1, 7, 10, 11¦, ¦2, 3, 5, 6¦, ¦4, 8¦ and ¦9¦. Figure 4.5(b) shows their array
representation. Clearly, since the elements are consecutive integers, the ar-
ray representation is preferable. However, in developing the algorithms for
the union and ﬁnd operations, we will assume the more general represen-
tation, that is, the tree representation.
Disjoint Sets Data Structures 127
10 1 2 3 4 5 6 7 8 9 11
0 3 0 8 2 2 1 0 0 1 1
(b)
1
7 10 11
3
2
5 6
8
4 9
(a)
Fig. 4.5 An example of the representation of disjoint sets. (a) Tree representation.
(b) Array representation when S = {1, 2, . . . , n}.
Now that we have deﬁned the data structure, we focus our attention on
the implementation of the union and ﬁnd operations. A straightforward
implementation is as follows. In the case of the operation find(x), sim-
ply follow the path from x until the root is reached, then return root(x).
In the case of the operation union(x, y), let the link of root(x) point to
root(y), i.e., if root(x) is u and root(y) is v, then let v be the parent of u.
In order to improve on the running time, we present in the following
two sections two heuristics: union by rank and path compression.
4.3.1 The union by rank heuristic
An obvious disadvantage of the straightforward implementation of the
union operation stated above is that the height of a tree may become very
large to the extent that a ﬁnd operation may require Ω(n) time. In the ex-
treme case, a tree may become degenerate. A simple example of this case
is in order. Suppose we start with the singleton sets ¦1¦, ¦2¦, . . . , ¦n¦ and
then execute the following sequence of unions and ﬁnds (see Fig. 4.6(a)):
union(1, 2), union(2, 3), . . . , union(n −1, n),
find(1), find(2), . . . , find(n).
128 Heaps and the Disjoint Sets Data Structures
n
¬
·
1 n-1
(b)
n
n-1
¬
1
(a)
Fig. 4.6 The result of n − 1 union operations. (a) Without union by rank. (b) With
union by rank.
In this case, the total cost of the n ﬁnd operations is proportional to
n + (n −1) +. . . + 1 =
n(n + 1)
2
= Θ(n
2
).
In order to constraint the height of each tree, we adopt the union by rank
heuristic: We store with each node a nonnegative number referred to as
the rank of that node. The rank of a node is essentially its height. Let
x and y be two roots of two diﬀerent trees in the current forest. Initially,
each node has rank 0. When performing the operation union(x, y), we
compare rank(x) and rank(y). If rank(x) < rank(y), we make y the parent
of x. If rank(x) > rank(y), we make x the parent of y. Otherwise, if
rank(x) = rank(y), we make y the parent of x and increase rank(y) by
one. Applying this rule on the sequence of operations above yields the tree
shown in Fig. 4.6(b). Note that the total cost of the n ﬁnd operations is now
reduced to Θ(n). This, however, is not always the case. As will be shown
later, using this rule, the time required to process n ﬁnds is O(nlog n).
Let x be any node, and p(x) the parent of x. The following two obser-
vations are fundamental.
Observation 4.1 rank(p(x)) ≥ rank(x) + 1.
Observation 4.2 The value of rank(x) is initially zero and increases in
subsequent union operations until x is no longer a root. Once x becomes a
child of another node, its rank never changes.
Lemma 4.1 The number of nodes in a tree with root x is at least 2
rank(x)
.
Disjoint Sets Data Structures 129
Proof. By induction on the number of union operations. Initially, x is a
tree by itself and its rank is zero. Let x and y be two roots, and consider the
operation union(x, y). Assume that the lemma holds before this operation.
If rank(x) < rank(y), then the formed tree rooted at y has more nodes than
the old tree with root y and its rank is unchanged. If rank(x) > rank(y),
then the formed tree rooted at x has more nodes than the old tree with root
x and its rank is unchanged. Thus, if rank(x) = rank(y), then the lemma
holds after the operation. If, however, rank(x) = rank(y), then in this case,
by induction, the formed tree with root y has at least 2
rank(x)
+2
rank(y)
=
2
rank(y)+1
nodes. Since rank(y) will be increased by one, the lemma holds
after the operation. ⁄
Clearly, if x is the root of tree T, then the height of T is exactly the rank
of x. By Lemma 4.1, if the number of nodes in the tree rooted at x is k, then
the height of that tree is at most log k|. It follows that the cost of each
ﬁnd operation is O(log n). The time required by the operation union(x, y)
is O(1) if both arguments are roots. If not both x and y are roots, then
the running time reduces to that of the ﬁnd operation. Consequently, the
time complexity of a union operation is that of the ﬁnd operation, which
is O(log n). It follows that, using the union by rank heuristic, the time
complexity of a sequence of m interspersed union and ﬁnd instructions is
O(mlog n).
4.3.2 Path compression
To enhance the performance of the ﬁnd operation further, another heuristic
known as path compression is also employed. In a find(x) operation, after
the root y is found, we traverse the path from x to y one more time and
change the parent pointers of all nodes along the path to point directly to
y. The action of path compression is illustrated in Fig. 4.7.
During the execution of the operation find(4), the name of the set is
found to be 1. Therefore, the parent pointer of each node on the path from 4
to 1 is reset so that it points to 1. It is true that path compression increases
the amount of work required to perform a ﬁnd operation. However, this
process will pay oﬀ in subsequent ﬁnd operations, as we will be traversing
shorter paths. Note that when path compression is used, the rank of a node
may be greater than its height, so it serves as an upper bound on the height
of that node.
130 Heaps and the Disjoint Sets Data Structures
1
3 2
4
1
3
2
4
(a) (b)
Fig. 4.7 The eﬀect of executing the ﬁnd operation find(4) with path compression.
4.3.3 The union-ﬁnd algorithms
Algorithms find and union describe the ﬁnal versions of the ﬁnd and union
operations using the two heuristics stated above.
Algorithm 4.6 find
Input: A node x
Output: root(x), the root of the tree containing x.
1. y←x
2. while p(y) ,= null ¦Find the root of the tree containing x¦
3. y←p(y)
4. end while
5. root ←y; y←x
6. while p(y) ,= null ¦Do path compression¦
7. w←p(y)
8. p(y)←root
9. y←w
10. end while
11. return root
Example 4.4 Let S = ¦1, 2, . . . , 9¦ and consider applying the following se-
quence of unions and ﬁnds: union(1, 2), union(3, 4), union(5, 6), union(7, 8),
union(2, 4), union(8, 9), union(6, 8), find(5), union(4, 8), find(1). Figure 4.8(a)
shows the initial conﬁguration. Figure 4.8(b) shows the data structure after the
ﬁrst four union operations. The result of the next three union operations is shown
Disjoint Sets Data Structures 131
Algorithm 4.7 union
Input: Two elements x and y
Output: The union of the two trees containing x and y. The original trees
are destroyed.
1. u←find(x); v←find(y)
2. if rank(u) ≤ rank(v) then
3. p(u)←v
4. if rank(u) = rank(v) then rank(v)←rank(v) + 1
5. else p(v)←u
6. end if
in Fig. 4.8(c). Figure 4.8(d) shows the eﬀect of executing the operation find(5).
The results of the operations union(4, 8) and find(1) are shown in Figs. 4.8(e)
and 4.8(f), respectively.
7
8
9
5
6
1
2
3
(f)
7
8
9
5
6
3
4
1
2
(e)
3
4
1
2
(c)
7
8
9
3
4
1
2
(d)
7
8
9
5
6
9
(b)
3
4
1
2
5
6
7
8
1 2 3 4 5
6 7 8 9
(a)
4
5
6
Fig. 4.8 An example of the union-ﬁnd algorithms.
132 Heaps and the Disjoint Sets Data Structures
4.3.4 Analysis of the union-ﬁnd algorithms
We have shown before that the worst case running time required to process
an interspersed sequence σ of m union and ﬁnd operations using union by
rank is O(mlog n). Now we show that if path compression is also employed,
then using amortized time analysis (see Sec. 1.13), it is possible to prove
that the bound is almost O(m).
Lemma 4.2 For any integer r ≥ 0, the number of nodes of rank r is at
most n/2
r
.
Proof. Fix a particular value of r. When a node x is assigned a rank of
r, label by x all the nodes contained in the tree rooted at x. By Lemma 4.1,
the number of labeled nodes is at least 2
r
. If the root of that tree changes,
then the rank of the root of the new tree is at least r +1. This means that
those nodes labeled with x will never be labeled again. Since the maximum
number of nodes labeled is n, and since each root of rank r has at least 2
r
nodes, it follows that there are at most n/2
r
nodes with rank r. ⁄
Corollary 4.1 The rank of any node is at most log n|.
Proof. If for some node x, rank(x) = r ≥ log n|+1, then by Lemma 4.2,
there are at most n/2
log n+1
< 1 nodes of rank r. ⁄
Deﬁnition 4.2 For any positive integer n, log
∗
n is deﬁned as
log
∗
n =

1 if j = 0
2
F(j−1)
if j ≥ 1.
The most important property of F(j) is its explosive growth. For example,
F(1) = 2, F(2) = 4, F(3) = 16, F(4) = 65536 and F(5) = 2
65536
.
Let σ be a sequence of m union and ﬁnd instructions. We partition the
ranks into groups. We put rank r in group log
∗
r. For example, ranks 0
and 1 are in group 0, rank 2 is in group 1, ranks 3 and 4 are in group 2,
Disjoint Sets Data Structures 133
ranks 5 through 16 are in group 3 and ranks 17 through 65536 are in group
4. Since the largest possible rank is log n|, the largest group number is at
most log
∗
log n = log
∗
n −1.
We assess the charges of a ﬁnd instruction find(u) as follows. Let v be
a node on the path from node u to the root of the tree containing u, and
let x be that root. If v is the root, a child of the root, or if the parent of
v is in a diﬀerent rank group from v, then charge one time unit to the ﬁnd
instruction itself. If v = x, and both v and its parent are in the same rank
group, then charge one time unit to node v. Note that the nodes on the
path from u to x are monotonically increasing in rank, and since there are
at most log
∗
n diﬀerent rank groups, no ﬁnd instruction is charged more
than O(log
∗
n) time units. It follows that the total number of time units
charged to all the ﬁnd instructions in the sequence σ is O(mlog
∗
n).
After x is found to be the root of the tree containing u, then by applying
path compression, x will be the parent of both u and v. If later on x becomes
a child of another node, and v and x are in diﬀerent groups, no more node
costs will be charged to v in subsequent ﬁnd instructions. An important
observation is that if node v is in rank group g > 0, then v can be moved
and charged at most F(g) − F(g − 1) times before it acquires a parent in
a higher group. If node v is in rank group 0, it will be moved at most once
before having a parent in a higher group.
Now we derive an upper bound on the total charges made to the nodes.
By Lemma 4.2, the number of nodes of rank r is at most n/2
r
. If we deﬁne
F(−1) = 0, then the number of nodes in group g is at most
F(g)
¸
r=F(g−1)+1
n
2
r
≤
n
2
F(g−1)+1
∞
¸
r=0
1
2
r
=
n
2
F(g−1)
=
n
F(g)
.
Since the maximum number of node charges assigned to a node in group
g is equal to F(g) − F(g − 1), the number of node charges assigned to all
134 Heaps and the Disjoint Sets Data Structures
nodes in group g is at most
n
F(g)
(F(g) −F(g −1)) ≤ n.
Since there are at most log
∗
n groups (0, 1, . . . , log
∗
n−1), it follows that the
number of node charges assigned to all nodes is O(nlog
∗
n). Combining this
with the O(mlog
∗
n) charges to the ﬁnd instructions yields the following
theorem:
Theorem 4.3 Let T(m) denote the running time required to process an
interspersed sequence σ of m union and ﬁnd operations using union by rank
and path compression. Then T(m) = O(mlog
∗
n) in the worst case.
Note that for almost all practical purposes, log
∗
n ≤ 5. This means that
the running time is O(m) for virtually all practical applications.
4.4 Exercises
4.1. What are the merits and demerits of implementing a priority queue using
an ordered list?
4.2. What are the costs of insert and delete-max operations of a priority
queue that is implemented as a regular queue.
4.3. Which of the following arrays are heaps?
(a) 8 6 4 3 2 . (b) 7 . (c) 9 7 5 6 3 .
(d) 9 4 8 3 2 5 7 . (e) 9 4 7 2 1 6 5 3 .
4.4. Where do the following element keys reside in a heap?
(a) Second largest key. (b) Third largest key. (c) Minimum key.
4.5. Give an eﬃcient algorithm to test whether a given array A[1..n] is a
heap. What is the time complexity of your algorithm?
4.6. Which heap operation is more costly: insertion or deletion? Justify your
answer. Recall that both operations have the same time complexity, that
is, O(log n).
4.7. Let H be the heap shown in Fig. 4.1. Show the heap that results from
(a) deleting the element with key 17.
(b) inserting an element with key 19.
4.8. Show the heap (in both tree and array representation) that results from
deleting the maximum key in the heap shown in Fig. 4.4(e).
Exercises 135
4.9. How fast is it possible to ﬁnd the minimum key in a max-heap of n
elements?
4.10. Prove or disprove the following claim. Let x and y be two elements in a
heap whose keys are positive integers, and let T be the tree representing
that heap. Let h
x
and h
y
be the heights of x and y in T. Then, if x is
greater than y, h
x
cannot be less than h
y
. (See Sec. 3.5 for the deﬁnition
of node height).
4.11. Illustrate the operation of Algorithm makeheap on the array
3 7 2 1 9 8 6 4 .
4.12. Show the steps of transforming the following array into a heap
1 4 3 2 5 7 6 8 .
4.13. Let A[1..19] be an array of 19 integers, and suppose we apply Algorithm
makeheap on this array.
(a) How many calls to Procedure sift-down will there be? Explain.
(b) What is the maximum number of element interchanges in this case?
Explain.
(c) Give an array of 19 elements that requires the above maximum
number of element interchanges.
4.14. Show how to use Algorithm heapsort to arrange in increasing order the
integers in the array
4 5 2 9 8 7 1 3 .
4.15. Given an array A[1..n] of integers, we can create a heap B[1..n] from A as
follows. Starting from the empty heap, repeatedly insert the elements of
A into B, each time adjusting the current heap, until B contains all the
elements in A. Show that the running time of this algorithm is Θ(nlog n)
in the worst case.
4.16. Illustrate the operation of the algorithm in Exercise 4.15 on the array
6 9 2 7 1 8 4 3 .
4.17. Explain the behavior of Algorithm heapsort when the input array is
already sorted in
(a) increasing order.
(b) decreasing order.
4.18. Give an example of a binary search tree with the heap property.
136 Heaps and the Disjoint Sets Data Structures
4.19. Give an algorithm to merge two heaps of the same size into one heap.
What is the time complexity of your algorithm?
4.20. Compute the minimum and maximum number of element comparisons
performed by Algorithm heapsort.
4.21. A d-heap is a generalization of the binary heap discussed in this chapter.
It is represented by an almost-complete d-ary rooted tree for some d ≥ 2.
Rewrite Procedure sift-up for the case of d-heaps. What is its time
complexity?
4.22. Rewrite Procedure sift-down for the case of d-heaps (see Exercise 4.21).
What is its time complexity measured in terms of d and n?
4.23. Give a sequence of n union and ﬁnd operations that results in a tree of
height Θ(log n) using only the heuristic of union by rank. Assume the
set of elements is ¦1, 2, . . . , n¦.
4.24. Give a sequence of n union and ﬁnd operations that requires Θ(nlog n)
time using only the heuristic of union by rank. Assume the set of ele-
ments is ¦1, 2, . . . , n¦.
4.25. What are the ranks of nodes 3, 4 and 8 in Fig. 4.8(f)?
4.26. Let ¦1¦, ¦2¦, ¦3¦, . . . , ¦8¦ be n singleton sets, each represented by a
tree with exactly one node. Use the union-ﬁnd algorithms with union
by rank and path compression to ﬁnd the tree representation of the
set resulting from each of the following unions and ﬁnds: union(1, 2),
union(3, 4),union(5, 6), union(7, 8), union(1, 3), union(5, 7), find(1),
union(1, 5), find(1).
4.27. Let T be a tree resulting from a sequence of unions and ﬁnds using both
the heuristics of union by rank and path compression, and let x be a
node in T. Prove that rank(x) is an upper bound on the height of x.
4.28. Let σ be a sequence of union and ﬁnd instructions in which all the unions
occur before the ﬁnds. Show that the running time is linear if both the
heuristics of union by rank and path compression are used.
4.29. Another heuristic that is similar to union by rank is the weight-balancing
rule. In this heuristic, the action of the operation union(x, y) is to let
the root of the tree with fewer nodes point to the root of the tree with
a larger number of nodes. If both trees have the same number of nodes,
then let y be the parent of x. Compare this heuristic with the union by
rank heuristic.
4.30. Solve Exercise 4.26 using the weight-balancing rule and path compression
(see Exercise 4.29).
4.31. Prove that the weight-balancing rule described in Exercise 4.29 guaran-
tees that the resulting tree is of height O(log n).
4.32. Let T be a tree resulting from a sequence of unions and ﬁnds using the
Bibliographic notes 137
heuristics of union by rank and path compression. Let x be the root of
T and y a leaf node in T. Prove that the ranks of the nodes on the path
from y to x form a strictly increasing sequence.
4.33. Prove the observation that if node v is in rank group g > 0, then v can
be moved and charged at most F(g) −F(g −1) times before it acquires
a parent in a higher group.
4.34. Another possibility for the representation of disjoint sets is by using
linked lists. Each set is represented by a linked list, where the set rep-
resentative is the ﬁrst element in the list. Each element in the list has
a pointer to the set representative. Initially, one list is created for each
element. The union of two sets is implemented by merging the two sets.
Suppose two sets S
1
represented by list L
1
and S
2
represented by list L
2
are to be merged. If the ﬁrst element in L
1
is to be used as the name of
the resulting set, then the pointer to the set name at each element in L
2
must be changed so that it points to the ﬁrst element in L
1
.
(a) Explain how to improve this representation so that each ﬁnd oper-
ation takes O(1) time.
(b) Show that the total cost of performing n−1 unions is Θ(n
2
) in the
worst case.
4.35. (Refer to Exercise 4.34). Show that if when performing the union of
two sets, the ﬁrst element in the list with a larger number of elements
is always chosen as the name of the new set, then the total cost of
performing n −1 unions becomes O(nlog n).
4.5 Bibliographic notes
Heaps and the data structures for disjoint sets appear in several books on
algorithms and data structures (see the bibliographic notes of chapters 1
and 3). They are covered in greater depth in Tarjan (1983). Heaps were
ﬁrst introduced as part of heapsort by Williams (1964). The linear time al-
gorithm for building a heap is due to Floyd (1964). A number of variants of
heaps can be found in Cormen et al. (1992), e.g. binomial heaps, Fibonacci
heaps. A comparative study of many data structures for priority queues can
be found in Jones (1986). The disjoint sets data structure was ﬁrst studied
by Galler and Fischer (1964) and Fischer (1972). A more detailed analysis
was carried out by Hopcroft and Ullman (1973) and then a more exact
analysis by Tarjan (1975). In this paper, a lower bound that is not linear
was established when both union by rank and path compression are used.
138
PART 2
Techniques Based on Recursion
139
140
141
This part of the book is concerned with a particular class of algorithms,
called recursive algorithms. These algorithms turn out to be of fundamental
importance and indispensible in virtually every area in the ﬁeld of computer
science. For many problems, the use of recursion makes it possible to solve
complex problems using algorithms that are concise, easy to comprehend
and eﬃcient (from an algorithmic point of view). In its simplest form, re-
cursion is the process of dividing the problem into one or more subproblems,
which are identical in structure to the original problem and then combin-
ing the solutions to these subproblems to obtain the solution to the original
problem. We identify three special cases of this general design technique:
(1) Induction or tail-recursion. (2) Nonoverlapping subproblems. (3) Over-
lapping subproblems with redundant invocations to subproblems, allowing
trading space for time. The higher numbered cases subsume the lower
numbered ones. The ﬁrst two cases will not require additional space for
the maintenance of solutions for continued reuse. The third class, however,
renders the possibility of eﬃcient solutions for many problems that at ﬁrst
glance appear to be time-consuming to solve.
Chapter 5 is devoted to the study of induction as a technique for the
development of algorithms. In other words, the idea of induction in math-
ematical proofs is carried over to the design of eﬃcient algorithms. In this
chapter, several examples are presented to show how to use induction to
solve increasingly sophisticated problems. These problems range from se-
lectionsort and insertionsort covered in Chapter 1 to the problems of
generating permutations and ﬁnding the majority in a multiset.
Chapter 6 provides a general overview of one of the most important
algorithm design techniques, namely divide and conquer. First, we de-
rive divide-and-conquer algorithms for the search problem and sorting by
merging. In particular, Algorithm mergesort is compared with Algorithm
bottomupsort presented in Chapter 1, which is an iterative version of the
former. This comparison reveals the most appealing merits of divide-and-
conquer algorithms: conciseness, ease of comprehension and implementa-
tion, and most importantly the simple inductive proofs for the correctness
of divide-and-conquer algorithms. Next, some useful algorithms like Al-
gorithm quicksort and Algorithm select for ﬁnding the kth smallest
element are discussed in detail.
Chapter 7 provides some examples of the use of dynamic programming
to solve problems for which recursion results in many redundant calls. In
this design technique, recursion is ﬁrst used to model the solution of the
142
problem. This recursive model is then converted into an eﬃcient iterative
algorithm. By trading space for time, this is achieved by saving results
of subproblems as they get solved and using a kind of table lookup for
future reference to those solved subproblems. In this chapter, dynamic
programming is applied to solve the longest common subsequence problem,
matrix chain multiplication, the all-pairs shortest path problem, and the
knapsack problem.
Chapter 5
Induction
5.1 Introduction
Consider a problem with parameter n, which normally represents the num-
ber of objects in an instance of the problem. When searching for a solution
to such a problem, sometimes it is easier to start with a solution to the
problem with a smaller parameter, say n −1, n/2, etc., and extend the so-
lution to include all the n objects. This approach is based on the well-known
proof technique of mathematical induction. Basically, given a problem with
parameter n, designing an algorithm by induction is based on the fact that
if we know how to solve the problem when presented with a parameter less
than n, called the induction hypothesis, then our task reduces to extending
that solution to include those instances with parameter n.
This method can be generalized to encompass all recursive algorithm
design techniques including divide and conquerand dynamic programming.
However, since these two have distinct marking characteristics, we will con-
ﬁne our attention in this chapter to those strategies that very much resemble
mathematical induction, and devote Chapters 6 and 7 to the study of divide
and conquer and dynamic programming, respectively. In the algorithm that
we will cover in this chapter, the resulting algorithms are usually recursive
with only one recursive call, commonly called tail recursion. Thus, in most
cases, they can conveniently be converted into iterative algorithms.
An advantage of this design technique (and all recursive algorithms
in general) is that the proof of correctness of the designed algorithm is
naturally embedded in its description, and a simple inductive proof can
easily be constructed if desired.
143
144 Induction
5.2 Two Simple Examples
In this section, we give two examples that reveal the essence of the induc-
tion technique. Speciﬁcally, we show how Algorithm selectionsort and
Algorithm insertionsort have simple derivations using induction.
5.2.1 Selection sort
In Sec. 1.5, we have presented Algorithm selectionsort. Now, we show
how to derive this algorithm using induction. Let the input to the algo-
rithm be A[1..n]. For the induction hypothesis, suppose we know how to
sort the last n − 1 elements, i.e., A[2..n]. Let the minimum be A[j], for
some j, 1 ≤ j ≤ n. First, we interchange A[1] and A[j] if j = 1. Next,
since by induction we know how to sort A[2..n], we recursively sort the ele-
ments in A[2..n]. This inductive reasoning leads to the recursive version of
Algorithm selectionsort shown as Algorithm selectionsortrec. Con-
verting this algorithm to an iterative one is straightforward (see Algorithm
selectionsort in Sec. 1.5).
Algorithm 5.1 selectionsortrec
Input: An array A[1..n] of n elements.
Output: A[1..n] sorted in nondecreasing order.
1. sort(1)
Procedure sort(i) ¦Sort A[i..n]¦
1. if i < n then
2. k←i
3. for j ←i + 1 to n
4. if A[j] < A[k] then k←j
5. end for
6. if k ,= i then interchange A[i] and A[k]
7. sort(i + 1)
8. end if
We compute the total number of element comparisons performed by
Algorithm selectionsortrec as follows. Let C(n) denote the number of
element comparisons performed by Procedure sort on an input of n ele-
ments, that is, when the array contains n elements. Clearly, C(1) = 0. The
ﬁrst time Procedure sort is invoked i = 1, and hence there are n elements
Radix Sort 145
in the array. Therefore, in this call, there are n − 1 element comparisons
plus the number of element comparisons that result from the recursive call
sort(i +1) to sort the array A[2..n]. Hence, the number of element compar-
isons that result from the recursive call is C(n−1). Putting these together,
we have
C(n) =

0 if n = 1
C(n −1) + (n −1) if n ≥ 2.
The solution to this recurrence is
C(n) =
n−1
¸
j=1
j =
n(n −1)
2
.
Since the running time is proportional to the number of element compar-
isons, we conclude that it is Θ(n
2
).
5.2.2 Insertion sort
In Sec. 1.6, we have also presented Algorithm insertionsort, an algo-
rithm that sorts an array of n elements in nondecreasing order. We now
show how to use induction technique to derive this algorithm. Let the input
to the algorithm be A[1..n]. Suppose we know how to sort the ﬁrst n − 1
elements, i.e., A[1..n −1]. Then after sorting A[1..n −1], we simply insert
A[n] in its proper position, say j, 1 ≤ j ≤ n. This insertion may result in
shifting A[j +1], A[j +2], . . . , A[n −1] to positions j +2, j +3, . . . , n. This
inductive reasoning leads to the recursive version of Algorithm insertion-
sort shown as Algorithm insertionsortrec. Converting this algorithm
to an iterative one is straightforward (see Algorithm insertionsort in
Sec. 1.6).
5.3 Radix Sort
In this section, we study a sorting algorithm that runs in linear time in al-
most all practical purposes. Let L = ¦a
1
, a
2
, . . . , a
n
¦ be a list of n numbers
each consisting of exactly k digits. That is, each number is of the form
d
k
d
k−1
. . . d
1
, where each d
i
is a digit between 0 and 9. In this problem,
instead of applying induction on n, the number of objects, we use induc-
tion on k, the size of each integer. One way to sort the numbers in L is to
146 Induction
Algorithm 5.2 insertionsortrec
Input: An array A[1..n] of n elements.
Output: A[1..n] sorted in nondecreasing order.
1. sort(n)
Procedure sort(i) ¦Sort A[1..i]¦
1. if i > 1 then
2. x←A[i]
3. sort(i −1)
4. j ←i −1
5. while j > 0 and A[j] > x
6. A[j + 1] ←A[j]
7. j ←j −1
8. end while
9. A[j + 1] ←x
10. end if
distribute them into 10 lists L
0
, L
1
, . . . , L
9
by their most signiﬁcant digit,
so that those numbers with d
k
= 0 constitute list L
0
, those with d
k
= 1
constitute list L
1
and so on. At the end of this step, for each i, 0 ≤ i ≤ 9,
L
i
contains those numbers whose most signiﬁcant digit is i. We have two
choices now. The ﬁrst choice is to sort each list using another sorting algo-
rithm and then concatenate the resulting lists into one sorted list. Observe
that in the worst case all numbers may have the same most signiﬁcant digit,
which means that they will all end up in one list and the other nine lists
will be empty. Hence, if the sorting algorithm used runs in Θ(nlog n) time,
the running time of this method will be Θ(nlog n). Another possibility is
to recursively sort each list on digit d
k−1
. But this approach will result in
the addition of more and more lists, which is undesirable.
Surprisingly, it turns out that if the numbers are ﬁrst distributed into the
lists by their least signiﬁcant digit, then a very eﬃcient algorithm results.
This algorithm is commonly known as radix sort. It is straightforward
to derive the algorithm using induction on k. Suppose that the numbers
are sorted lexicographically according to their least k −1 digits, i.e., digits
d
k−1
, d
k−2
, . . . , d
1
. Then, after sorting them on their kth digits, they will
eventually be sorted. The implementation of the algorithm does not require
any other sorting algorithm. Nor does it require recursion. The algorithm
works as follows. First, distribute the numbers into 10 lists L
0
, L
1
, . . . , L
9
according to digit d
1
, so that those numbers with d
1
= 0 constitute list
Radix Sort 147
L
0
, those with d
1
= 1 constitute list L
1
and so on. Next, the lists are
coalesced in the order L
0
, L
1
, . . . , L
9
. Then, they are distributed into 10
lists according to digit d
2
, coalesced in order, and so on. After distributing
them according to d
k
and collecting them in order, all numbers will be
sorted. The following example illustrates the idea.
Example 5.1 Figure 5.1 shows an example of radix sort. The left column
in the ﬁgure shows the input numbers. Successive columns show the results after
sorting by the 1st, 2nd, 3rd and 4th digit.
7467 6792 9134 9134 1239
1247 9134 1239 9187 1247
3275 3275 1247 1239 3275
6792 4675 7467 1247 4675
9187 7467 3275 3275 6792
9134 1247 4675 7467 7467
4675 9187 9187 4675 9134
1239 1239 6792 6792 9187
Fig. 5.1 Example of radix sort.
Algorithm 5.3 radixsort
Input: A linked list of numbers L = ¦a
1
, a
2
, . . . , a
n
¦ and k, the number of
digits.
Output: L sorted in nondecreasing order.
1. for j ←1 to k
2. Prepare 10 empty lists L
0
, L
1
, . . . , L
9
.
3. while L is not empty
4. a← next element in L. Delete a from L.
5. i ←jth digit in a. Append a to list L
i
.
6. end while
7. L←L
0
8. for i ←1 to 9
9. L←L, L
i
¦append list L
i
to L¦
10. end for
11. end for
12. return L
148 Induction
The method is described more precisely in Algorithm radixsort.
There are k passes, and each pass costs Θ(n) time. Thus, the running
time of the algorithm is Θ(kn). If k is constant, the running time is simply
Θ(n). The algorithm uses Θ(n) space, as there are 10 lists needed and the
overall size of the lists is Θ(n).
It should be noted that the algorithm can be generalized to any radix,
not just radix 10 as in the algorithm. For example, we can treat each four
bits as one digit and work on radix 16. The number of lists will always be
equal to the radix. More generally, we can use Algorithm radixsort to
sort whole records on each ﬁeld. If, for example, we have a ﬁle of dates each
consisting of year, month, and day, we can sort the whole ﬁle by sorting
ﬁrst by day, then by month and ﬁnally by year.
5.4 Integer Exponentiation
In this section, we develop an eﬃcient algorithm for raising a real number
x to the nth power, where n is a nonnegative integer. The straightforward
method is to iteratively multiply x by itself n times. This method is very
ineﬃcient, as it requires Θ(n) multiplications, which is exponential in the
input size (see Sec. 1.14). An eﬃcient method can be deduced as follows.
Let m = n/2|, and suppose we know how to compute x
m
. Then, we have
two cases: If n is even, then x
n
= (x
m
)
2
; otherwise x
n
= x(x
m
)
2
. This idea
immediately yields the recursive algorithm shown as Algorithm exprec.
Algorithm 5.4 exprec
Input: A real number x and a nonnegative integer n.
Output: x
n
.
1. power(x, n)
Procedure power(x, m) ¦Compute x
m
¦
1. if m = 0 then y←1
2. else
3. y←power(x, ]m/2)
4. y←y
2
5. if m is odd then y←xy
6. end if
7. return y
Evaluating Polynomials (Horner’s Rule) 149
Algorithm exprec can be rendered iterative using repeated squaring as
follows. Let the binary digits of n be d
k
, d
k−1
, . . . , d
0
. Starting from y = 1,
we scan the binary digits from left to right. If the current binary digit is
0, we simply square y and if it is 1, we square y and multiply it by x. This
yields Algorithm exp.
Algorithm 5.5 exp
Input: A real number x and a nonnegative integer n.
Output: x
n
.
1. y←1
2. Let n be d
k
d
k−1
. . . d
0
in binary notation.
3. for j ←k downto 0
4. y←y
2
5. if d
j
= 1 then y←xy
6. end for
7. return y
Assuming that each multiplication takes constant time, the running time
of both versions of the algorithm is Θ(log n), which is linear in the input size.
5.5 Evaluating Polynomials (Horner’s Rule)
Suppose we have a sequence of n+2 real numbers a
0
, a
1
, . . . , a
n
and x, and
we want to evaluate the polynomial
P
n
(x) = a
n
x
n
+a
n−1
x
n−1
+. . . +a
1
x +a
0
.
The straightforward approach would be to evaluate each term separately.
This approach is very ineﬃcient since it requires n+n−1 . . .+1 = n(n+1)/2
multiplications.
A much faster method can be derived by induction as follows. First, we
observe that
P
n
(x) = a
n
x
n
+a
n−1
x
n−1
+. . . +a
1
x +a
0
= ((. . . (((a
n
x +a
n−1
)x +a
n−2
)x +a
n−3
) . . .)x +a
1
)x +a
0
.
This evaluation scheme is known as Horner’s rule. Using this scheme leads
150 Induction
to the following more eﬃcient method. Suppose we know how to evaluate
P
n−1
(x) = a
n
x
n−1
+a
n−1
x
n−2
+. . . +a
2
x +a
1
.
Then, using one more multiplication and one more addition, we have
P
n
(x) = xP
n−1
(x) +a
0
.
This implies Algorithm horner.
Algorithm 5.6 horner
Input: A sequence of n + 2 real numbers a
0
, a
1
, . . . , a
n
and x.
Output: P
n
(x) = a
n
x
n
+a
n−1
x
n−1
+. . . +a
1
x +a
0
.
1. p←a
n
2. for j ←1 to n
3. p←xp +a
n−j
4. end for
5. return p
It is easy to see that Algorithm horner costs n multiplications and
n additions. This is a remarkable achievement, which is attributed to the
judicious choice of the induction hypothesis.
5.6 Generating Permutations
In this section, we study the problem of generating all permutations of the
numbers 1, 2, . . . , n. We will use an array P[1..n] to hold each permutation.
Using induction, it is fairly easy to derive several algorithms. In this section,
we will present two of them that are based on the assumption that we can
generate all the permutations of n −1 numbers.
5.6.1 The ﬁrst algorithm
Suppose we can generate all permutations of n −1 numbers. Then, we can
extend our method to generate the permutations of the numbers 1, 2, . . . , n
as follows. Generate all the permutations of the numbers 2, 3, . . . , n and
add the number 1 to the beginning of each permutation. Next, generate
all permutations of the numbers 1, 3, 4, . . . , n and add the number 2 to
the beginning of each permutation. Repeat this procedure until ﬁnally the
Generating Permutations 151
permutations of 1, 2, . . . , n−1 are generated and the number n is added at
the beginning of each permutation. This method is described in Algorithm
permutations1. Note that when P[j] and P[m] are interchanged before
the recursive call, they must be interchanged back after the recursive call.
Algorithm 5.7 permutations1
Input: A positive integer n.
Output: All permutations of the numbers 1, 2, . . . , n.
1. for j ←1 to n
2. P[j] ←j
3. end for
4. perm1(1)
Procedure perm1(m)
1. if m = n then output P[1..n]
2. else
3. for j ←m to n
4. interchange P[j] and P[m]
5. perm1(m+ 1)
6. interchange P[j] and P[m]
7. comment: At this point P[m..n] = m, m+1, . . . , n.
8. end for
9. end if
We analyze the running time of the algorithm as follows. Since there
are n! permutations, Step 1 of Procedure perm1 takes nn! to output all
permutations. Now we count the number of iterations of the for loop. In
the ﬁrst call to Procedure perm1, m = 1. Hence, the for loop is executed n
times plus the number of times it is executed in the recursive call perm1(2).
Since when n = 1 the number of iterations is zero, the number of iterations
f(n) can be expressed by the recurrence
f(n) =

0 if n = 1
nf(n −1) +n if n ≥ 2.
Following the technique outlined in Sec. 2.8.2, we proceed to solve this
recurrence as follows. Let n!h(n) = f(n) (note that h(1) = 0). Then,
n!h(n) = n(n −1)!h(n −1) +n,
152 Induction
or
h(n) = h(n −1) +
n
n!
= h(n −1) +
1
(n −1)!
.
The solution to this recurrence is
h(n) = h(1) +
n
¸
j=2
1
(j −1)!
=
n−1
¸
j=1
1
j!
<
∞
¸
j=1
1
j!
= (e −1),
where e = 2.7182818 . . . (Eq. 2.1). Hence,
f(n) = n!h(n) = n!
n−1
¸
j=1
1
j!
< 2n!.
Hence, the running time of the algorithm is dominated by the output state-
ment, that is, Θ(nn!).
5.6.2 The second algorithm
In this section, we describe another algorithm for enumerating all the per-
mutations of the numbers 1, 2, . . . , n. At the beginning, all n entries of the
array P[1..n] are free, and each free entry will be denoted by 0. For the
induction hypothesis, let us assume that we have a method that generates
all permutations of the numbers 1, 2, . . . , n − 1. Then, we can extend our
method to generate all the permutations of the n numbers as follows. First,
we put n in P[1] and generate all the permutations of the ﬁrst n −1 num-
bers using the subarray P[2..n]. Next, we put n in P[2] and generate all
the permutations of the ﬁrst n − 1 numbers using the subarrays P[1] and
P[3..n]. Then, we put n in P[3] and generate all the permutations of the
ﬁrst n−1 numbers using the subarrays P[1..2] and P[4..n]. This continues
until ﬁnally we put n in P[n] and generate all the permutations of the ﬁrst
n − 1 numbers using the subarray P[1..n − 1]. Initially, all n entries of
P[1..n] contain 0’s. The method is described more precisely in Algorithm
permutations2.
Algorithm permutations2 works as follows. If the value of m is equal
to 0, then this is an indication that Procedure perm2 has been called for
all consecutive values n, n −1, . . . , 1. In this case, array P[1..n] has no free
entries and contains a permutation of ¦1, 2, . . . , n¦. If, on the other hand,
m > 0, then we know that m+ 1, m+ 2, . . . , n have already been assigned
to some entries of the array P[1..n]. Thus, we search for a free entry P[j] in
Generating Permutations 153
Algorithm 5.8 permutations2
Input: A positive integer n.
Output: All permutations of the numbers 1, 2, . . . , n.
1. for j ←1 to n
2. P[j] ←0
3. end for
4. perm2(n)
Procedure perm2(m)
1. if m = 0 then output P[1..n]
2. else
3. for j ←1 to n
4. if P[j] = 0 then
5. P[j] ←m
6. perm(m−1)
7. P[j] ←0
8. end if
9. end for
10. end if
the array and set P[j] to m, and then we call Procedure perm2 recursively
with parameter m − 1. After the recursive call, we must set P[j] to 0
indicating that it is now free and can be used in subsequent calls.
We analyze the running time of the algorithm as follows. Since there
are n! permutations, Step 1 of Procedure perm2 takes nn! to output all
permutations. Now we count the number of iterations of the for loop. The
for loop is executed n times in every call perm2(m) plus the number of
times it is executed in the recursive call perm2(m− 1). When Procedure
perm2 is invoked by the call perm2(m) with m > 0, the array P contains
exactly m zeros, and hence the recursive call perm2(m−1) will be executed
exactly m times. Since when m = 0 the number of iterations is zero, the
number of iterations can be expressed by the recurrence
f(m) =

0 if m = 0
mf(m−1) +n if m ≥ 1.
It should be emphasized that n in the recurrence above is constant and
independent of the value of m.
Following the technique outlined in Sec. 2.8.2, we proceed to solve this
154 Induction
recurrence as follows. Let m!h(m) = f(m) (note that h(0) = 0). Then,
m!h(m) = m(m−1)!h(m−1) +n,
or
h(m) = h(m−1) +
n
m!
.
The solution to this recurrence is
h(m) = h(0) +
m
¸
j=1
n
j!
= n
m
¸
j=1
1
j!
<
∞
¸
j=1
1
j!
= (e −1),
where e = 2.7182818 . . . (Eq. 2.1).
Hence,
f(m) = m!h(m) = nm!
m
¸
j=1
1
j!
< 2nm!.
In terms of n, the number of iterations becomes
f(n) = nn!
n
¸
j=1
1
j!
< 2nn!.
Hence, the running time of the algorithm is Θ(nn!).
5.7 Finding the Majority Element
Let A[1..n] be a sequence of integers. An integer a in A is called the ma-
jority if it appears more than n/2| times in A. For example, 3 is the
majority in the sequence 1, 3, 2, 3, 3, 4, 3, since it appears four times
and the number of elements is seven. There are several ways to solve this
problem. The brute-force method is to compare each element with every
other element and produce a count for each element. If the count of some
element is more than n/2|, then it is declared as the majority; otherwise
there is no majority in the list. But the number of comparisons here is
n(n −1)/2 = Θ(n
2
), which makes this method too costly. A more eﬃcient
algorithm is to sort the elements and count how many times each element
appears in the sequence. This costs Θ(nlog n) in the worst case, as the sort-
ing step requires Ω(nlog n) comparisons in the worst case (Theorem 12.2).
Another alternative is to ﬁnd the median, i.e., the n/2|th element. Since
Exercises 155
the majority must be the median, we can scan the sequence to test if the
median is indeed the majority. This method takes Θ(n) time, as the median
can be found in Θ(n) time. As we will see in Sec. 6.5, the hidden constant
in the time complexity of the median ﬁnding algorithm is too large, and
the algorithm is fairly complex.
It turns out that there is an elegant solution that uses much fewer
comparisons. We derive this algorithm using induction. The essence of the
algorithm is based on the following observation:
Observation 5.1 If two diﬀerent elements in the original sequence are
removed, then the majority in the original sequence remains the majority
in the new sequence.
This observation suggests the following procedure for ﬁnding an element
that is a candidate for being the majority. Set a counter to zero and let
x = A[1]. Starting from A[2], scan the elements one by one increasing
the counter by one if the current element is equal to x and decreasing
the counter by one if the current element is not equal to x. If all the
elements have been scanned and the counter is greater than zero, then
return x as the candidate. If the counter becomes zero when comparing
x with A[j], 1 < j < n, then call procedure candidate recursively on the
elements A[j +1..n]. Notice that decrementing the counter implements the
idea of throwing two diﬀerent elements as stated in Observation 5.1. This
method is described more precisely in Algorithm majority. Converting
this recursive algorithm into an iterative one is straightforward, and is left
as an exercise.
5.8 Exercises
5.1. Give a recursive algorithm that computes the nth Fibonacci number f
n
deﬁned by
f
1
= f
2
= 1; f
n
= f
n−1
+f
n−2
for n ≥ 3.
5.2. Give an iterative algorithm that computes the nth Fibonacci number f
n
deﬁned above.
5.3. Use induction to develop a recursive algorithm for ﬁnding the maximum
element in a given sequence A[1..n] of n elements.
156 Induction
Algorithm 5.9 majority
Input: An array A[1..n] of n elements.
Output: The majority element if it exists; otherwise none.
1. c←candidate(1)
2. count ←0
3. for j ←1 to n
4. if A[j] = c then count ←count + 1
5. end for
6. if count > ]n/2 then return c
7. else return none
Procedure candidate(m)
1. j ←m; c←A[m]; count ←1
2. while j < n and count > 0
3. j ←j + 1
4. if A[j] = c then count ←count + 1
5. else count ←count −1
6. end while
7. if j = n then return c ¦See Exercises 5.31 and 5.32.¦
8. else return candidate(j + 1)
5.4. Use induction to develop a recursive algorithm for ﬁnding the average of
n real numbers A[1..n].
5.5. Use induction to develop a recursive algorithm that searches for an ele-
ment x in a given sequence A[1..n] of n elements.
5.6. Derive the running time of Algorithm insertionsortrec.
5.7. Illustrate the operation of Algorithm radixsort on the following se-
quence of eight numbers:
(a) 4567, 2463, 6523, 7461, 4251, 3241, 6492, 7563.
(b) 16543, 25895, 18674, 98256, 91428, 73234, 16597, 73195.
5.8. Express the time complexity of Algorithm radixsort in terms of n when
the input consists of n positive integers in the interval
(a) [1..n].
(b) [1..n
2
].
(c) [1..2
n
].
5.9. Let A[1..n] be an array of positive integers in the interval [1..n!]. Which
sorting algorithm do you think is faster: bottomupsort or radixsort?
(See Sec. 1.7).
5.10. What is the time complexity of Algorithm radixsort if arrays are used
instead of linked lists? Explain.
Exercises 157
5.11. Give a recursive version of Algorithm bubblesort given in Exercise 1.16.
5.12. A sorting method known as bucket sort works as follows. Let A[1..n]
be a sequence of n numbers within a reasonable range, say all numbers
are between 1 and m, where m is not too large compared to n. The
numbers are distributed into k buckets, with the ﬁrst bucket containing
those numbers between 1 and ]m/k, the second bucket containing those
numbers between ]m/k+1 to ]2m/k, and so on. The numbers in each
bucket are then sorted using another sorting algorithm, say Algorithm
insertionsort. Analyze the running time of the algorithm.
5.13. Instead of using another sorting algorithm in Exercies 5.12, design a
recursive version of bucket sort that recursively sorts the numbers in
each bucket. What is the major disadvantage of this recursive version?
5.14. A sorting algorithm is called stable if the order of equal elements is
preserved after sorting. Which of the following sorting algorithms are
stable?
(a)selectionsort (b)insertionsort (c)bubblesort
(d)bottomupsort (e)heapsort (f)radixsort.
5.15. Use induction to solve Exercise 3.7.
5.16. Use induction to solve Exercise 3.8.
5.17. Use Horner’s rule described in Sec. 5.5 to evaluate the following polyno-
mials:
(a) 3x
5
+ 2x
4
+ 4x
3
+x
2
+ 2x + 5.
(b) 2x
7
+ 3x
5
+ 2x
3
+ 5x
2
+ 3x + 7.
5.18. Use Algorithm exprec to compute
(a) 2
5
. (b) 2
7
. (c) 3
5
. (d) 5
7
.
5.19. Solve Exercise 5.18 using Algorithm exp instead of Algorithm exprec.
5.20. Carefully explain why in Algorithm permutations1 when P[j] and P[m]
are interchanged before the recursive call, they must be interchanged
back after the recursive call.
5.21. Carefully explain why in Algorithm permutations2 P[j] must be reset
to 0 after the recursive call.
5.22. Carefully explain why in Algorithm permutations2, when Procedure
perm2 is invoked by the call perm2(m) with m > 0, the array P con-
tains exactly m zeros, and hence the recursive call perm2(m−1) will be
executed exactly m times.
5.23. Modify Algorithm permutations2 so that the permutations of the num-
bers 1, 2, . . . , n are generated in a reverse order to that produced by
Algorithm permutations2.
158 Induction
5.24. Modify Algorithm permutations2 so that it generates all k-subsets of
the set ¦1, 2, . . . , n¦, 1 ≤ k ≤ n.
5.25. Analyze the time complexity of the modiﬁed algorithm in Exercise 5.24.
5.26. Prove the correctness of Algorithm permutations1.
5.27. Prove the correctness of Algorithm permutations2.
5.28. Give an iterative version of Algorithm majority.
5.29. Illustrate the operation of Algorithm majority on the arrays
(a) 5 7 5 4 5 .
(b) 5 7 5 4 8 .
(c) 2 4 1 4 4 4 6 4 .
5.30. Prove Observation 5.1.
5.31. Prove or disprove the following claim. If in Step 7 of Procedure candidate
in Algorithm majority j = n but count = 0 then c is the majority
element.
5.32. Prove or disprove the following claim. If in Step 7 of Procedure candidate
in Algorithm majority j = n and count > 0 then c is the majority
element.
5.33. Let A[1..n] be a sorted array of n integers, and x an integer. Design an
O(n) time algorithm to determine whether there are two elements in A,
if any, whose sum is exactly x.
5.9 Bibliographic notes
The use of induction as a mathematical technique for proving the correct-
ness of algorithms was ﬁrst developed by Floyd (1967). Recursion has been
studied extensively in algorithm design. See for example the books of Burge
(1975) and Paull (1988). The use of induction as a design technique appears
in Manber (1988). Manber (1989) is a whole book that is mostly devoted
to the induction design technique. Unlike this chapter, induction in that
book encompasses a wide variety of problems and is used in its broad sense
to cover other design techniques like divide and conquer and dynamic pro-
gramming. Radix sort is used by card-sorting machines. In old machines,
the machine did the distribution step and the operator collected the piles
after each pass and combined them into one for the next pass. Horner’s
rule for polynomial evaluation is after the English mathematician W. G.
Horner. Algorithm permutations2 appears in Banachowski, Kreczmar
Bibliographic notes 159
and Rytter (1991). The problem of ﬁnding the majority was studied for
example by Misra and Gries (1982). Fischer and Salzberg (1982) show that
using more sophisticated data structures, the number of comparisons can
be reduced to 3n/2 + 1 in the worst case, and this bound is optimal.
160
Chapter 6
Divide and Conquer
6.1 Introduction
The name “divide and conquer” has been given to a powerful algorithm
design technique that is used to solve a variety of problems. In its simplest
form, a divide-and-conquer algorithm divides the problem instance into a
number of subinstances (in most cases 2), recursively solves each subin-
stance separately, and then combines the solutions to the subinstances to
obtain the solution to the original problem instance. To illustrate this ap-
proach, consider the problem of ﬁnding both the minimum and maximum
in an array of integers A[1..n] and assume for simplicity that n is a power
of 2. A straightforward algorithm might look like the one below. It returns
a pair (x, y) where x is the minimum and y is the maximum.
1. x←A[1]; y←A[1]
2. for i ←2 to n
3. if A[i] < x then x←A[i]
4. if A[i] > y then y←A[i]
5. end for
6. return (x, y)
Clearly, the number of element comparisons performed by this method
is 2n−2. However, using the divide and conquer strategy, we can ﬁnd both
the minimum and maximum in only (3n/2) −2 element comparisons. The
idea is very simple: Divide the input array into two halves A[1..n/2] and
A[(n/2) + 1..n], ﬁnd the minimum and maximum in each half and return
161
162 Divide and Conquer
the minimum of the two minima and the maximum of the two maxima.
The divide-and-conquer algorithm is given in Algorithm minmax.
Algorithm 6.1 minmax
Input: An array A[1..n] of n integers, where n is a power of 2.
Output: (x, y):the minimum and maximum integers in A.
1. minmax(1, n)
Procedure minmax(low, high)
1. if high −low = 1 then
2. if A[low] < A[high] then return (A[low], A[high])
3. else return (A[high], A[low])
4. end if
5. else
6. mid ←](low + high)/2
7. (x
1
, y
1
)← minmax(low, mid)
8. (x
2
, y
2
)← minmax(mid + 1, high)
9. x← min¦x
1
, x
2
¦
10. y← max¦y
1
, y
2
¦
11. return (x, y)
12. end if
Let C(n) denote the number of comparisons performed by the algorithm
on an array of n elements, where n is a power of 2. Note that the element
comparisons are performed only in steps 2, 8 and 9. Also note that the
number of comparisons performed by steps 6 and 7 as a result of the recur-
sive calls is C(n/2). This gives rise to the following recurrence relation for
the number of comparisons done by the algorithm:
C(n) =

1 if n = 2
2C(n/2) + 2 if n > 2.
We proceed to solve this recurrence by expansion as follows (k = log n).
C(n) = 2C(n/2) + 2
= 2(2C(n/4) + 2) + 2
= 4C(n/4) + 4 + 2
= 4(2C(n/8) + 2) + 4 + 2
= 8C(n/8) + 8 + 4 + 2
.
.
.
Binary Search 163
= 2
k−1
C(n/2
k−1
) + 2
k−1
+ 2
k−2
+. . . + 2
2
+ 2
= 2
k−1
C(2) +
k−1
¸
j=1
2
j
= (n/2) + 2
k
−2 (Eq. 2.10, page 78)
= (3n/2) −2.
This result deserves to be called a theorem:
Theorem 6.1 Given an array A[1..n] of n elements, where n is a power
of 2, it is possible to ﬁnd both the minimum and maximum of the elements
in A using only (3n/2) −2 element comparisons.
6.2 Binary Search
Recall that in binary search, we test a given element x against the mid-
dle element in a sorted array A[low..high]. If x < A[mid], where mid =
(low + high)/2|, then we discard A[mid..high] and the same procedure is
repeated on A[low..mid − 1]. Similarly, if x > A[mid], then we discard
A[low..mid] and repeat the same procedure on A[mid +1..high]. This sug-
gests the recursive Algorithm binarysearchrec as another alternative to
implement this search method.
Algorithm 6.2 binarysearchrec
Input: An array A[1..n] of n elements sorted in nondecreasing order and
an element x.
Output: j if x = A[j], 1 ≤ j ≤ n, and 0 otherwise.
1. binarysearch(1, n)
Procedure binarysearch(low, high)
1. if low > high then return 0
2. else
3. mid ←](low + high)/2
4. if x = A[mid] then return mid
5. else if x < A[mid] then return binarysearch(low, mid −1)
6. else return binarysearch(mid + 1, high)
7. end if
164 Divide and Conquer
Analysis of the recursive binary search algorithm
To ﬁnd the running time of the algorithm, we compute the number of
element comparisons, since this is a basic operation, i.e., the running time
of the algorithm is proportional to the number of element comparisons
performed (see Sec. 1.11.2). We will assume that each three-way comparison
counts as one comparison. First, note that if n = 0, i.e., the array is empty,
then the algorithm does not perform any element comparisons. If n = 1, the
else part will be executed and, in case x = A[mid], the algorithm will recurse
on an empty array. It follows that if n = 1, then exactly one comparison is
performed. If n > 1, then there are two possibilities: If x = A[mid], then
only one comparison is performed; otherwise the number of comparisons
required by the algorithm is one plus the number of comparisons done
by the recursive call on either the ﬁrst or second half of the array. If
we let C(n) denote the number of comparisons performed by Algorithm
binarysearchrec in the worst case on an array of size n, then C(n) can
be expressed by the recurrence
C(n) ≤

1 if n = 1
1 +C(n/2|) if n ≥ 2.
Let k be such that 2
k−1
≤ n < 2
k
, for some integer k ≥ 2. If we expand
the above recurrence, we obtain
C(n) ≤ 1 +C(n/2|)
≤ 2 +C(n/4|)
.
.
.
≤ (k −1) +C(n/2
k−1
|)
= (k −1) + 1
= k,
since n/2|/2| = n/4| etc. (see Eq. 2.3 on page 71), and n/2
k−1
| = 1
(since 2
k−1
≤ n < 2
k
). Taking the logarithms of the inequalities
2
k−1
≤ n < 2
k
and adding 1 to both sides yields
k ≤ log n + 1 < k + 1,
Mergesort 165
or
k = log n| + 1,
since k is integer. It follows that
C(n) ≤ log n| + 1.
We have, in eﬀect, proved the following theorem:
Theorem 6.2 The number of element comparisons performed by Algo-
rithm binarysearchrec to search for an element in an array of n elements
is at most log n| + 1. Thus, the time complexity of Algorithm binary-
searchrec is O(log n).
We close this section by noting that the recursion depth is O(log n), and
since in each recursion level Θ(1) of space is needed, the total amount of
space needed by the algorithm is O(log n). Contrast this recursive algorithm
with the iterative version which needs only Θ(1) space (see Sec. 1.3).
6.3 Mergesort
In this section we consider an example of a simple divide-and-conquer algo-
rithm that reveals the essense of this algorithm design technique. We give
here more detailed description of how a generic divide-and-conquer algo-
rithm works in order to solve a problem instance in a top-down manner.
Consider the example of bottomupsort shown in Fig. 1.3. We have seen
how the elements were sorted by an implicit traversal of the associated sort-
ing tree level by level. In each level, we have pairs of sequences that have
already been sorted and are to be merged to obtain larger sorted sequences.
We continue ascending the tree level by level until we reach the root at
which the ﬁnal sequence has been sorted.
Now, let us consider doing the reverse, i.e., top-down instead of bottom-
up. In the beginning, we have the input array
A[1..8] = 9 4 5 2 1 7 4 6 .
We divide this array into the two 4-element arrays
9 4 5 2 and 1 7 4 6 .
166 Divide and Conquer
Next, we sort these two arrays individually, and then simply merge them
to obtain the desired sorted array. Call this algorithm sort. As to the
sorting method used for each half, we are free to make use of any sorting
algorithm to sort the two subarrays. In particular, we may use Algorithm
sort itself. If we do so, then we have indeed arrived at the well-known
mergesort algorithm. A precise description of this algorithm is given in
Algorithm mergesort.
Algorithm 6.3 mergesort
Input: An array A[1..n] of n elements.
Output: A[1..n] sorted in nondecreasing order.
1. mergesort(A, 1, n)
Procedure mergesort(low, high)
1. if low < high then
2. mid ←](low + high)/2
3. mergesort(A, low, mid)
4. mergesort(A, mid + 1, high)
5. merge (A, low, mid, high)
6. end if
A simple proof by induction establishes the correctness of the algorithm.
6.3.1 How the algorithm works
Consider Fig. 6.1, which illustrates the behavior of Algorithm mergesort
on the input array
A[1..8] = 9 4 5 2 1 7 4 6 .
As shown in the ﬁgure, the main call mergesort(A, 1, 8) induces a series
of recursive calls represented by an implicit binary tree. Each node of
the tree consists of two arrays. The top array is the input to the call
represented by that node, whereas the bottom array is its output. Each
edge of the tree is replaced with two antidirectional edges indicating the
ﬂow of control. The main call causes the call mergesort(A, 1, 4) to take
eﬀect, which, in turn, produces the call mergesort(A, 1, 2), and so on. Edge
labels indicate the order in which these recursive calls take place. This chain
of calls corresponds to a preorder traversal of the tree: Visit the root, the
Mergesort 167
6 4 7 1 2 5 9 4
6 4 7 1 2 5 9 4
2 5 1 6 4 9 4 7
5 2 1 6 4 4 9 7
4 2 5 9 1 4 7 6
4 9 5 2 1 6 4 7
2 9 7 6 4 5 1 4
4 1 6 9 7 4 5 2
3
4
5
6
9 10
11
12
17
18
19
20
23
24
25
26
8
2
7
13
16
21 22
27
15
1
14
28
Fig. 6.1 The behavior of Algorithm mergesort.
left subtree and then the right subtree (see Sec. 3.5.1). The computation,
however, corresponds to a postorder traversal of the tree: Visit the left
subtree, the right subtree and then the root. To implement this traversal,
a stack is used to hold the local data of each active call.
As indicated in the ﬁgure, the process of sorting the original array re-
duces to that used in Algorithm bottomupsort when n is a power of
2. Each pair of numbers are merged to produce two-element sorted se-
quences. These sorted 2-element sequences are then merged to obtain 4-
element sorted sequences, and so on. Compare Fig. 6.1 with Fig. 1.3. The
only diﬀerence between the two algorithms is in the order of merges: In
Algorithm bottomupsort, merges are carried out level by level, while
in Algorithm mergesort, the merges are performed in postorder. This
justiﬁes the remark stated in Observation 6.1 below that the number of
comparisons performed by mergesort is identical to that of Algorithm
bottomupsort when n is a power of 2.
6.3.2 Analysis of the mergesort algorithm
As in the binary search algorithm, the basic operation here is element com-
parison. That is, the running time is proportional to the number of element
comparisons performed by the algorithm. Now, we wish to compute the
number of element comparisons C(n) required by Algorithm mergesort
168 Divide and Conquer
to sort an array of n elements. For simplicity, we will assume that n is a
power of 2, i.e., n = 2
k
for some integer k ≥ 0. If n = 1, then the algorithm
does not perform any element comparisons. If n > 1, then steps 2 through
5 are executed. By deﬁnition of the function C, the number of compar-
isons required to execute steps 3 and 4 is C(n/2) each. By Observation 1.1,
the number of comparisons needed to merge the two subarrays is between
n/2 and n − 1. Thus, the minimum number of comparisons done by the
algorithm is given by the recurrence
C(n) =

0 if n = 1
2C(n/2) +n −1 if n ≥ 2
We proceed to solve this recurrence by expansion as follows
C(n) = 2C(n/2) +n −1
= 2(2C(n/2
2
) +n/2 −1) +n −1
= 2
2
C(n/2
2
) +n −2 +n −1
= 2
2
C(n/2
2
) + 2n −2 −1
.
.
.
= 2
k
C(n/2
k
) +kn −2
k−1
−2
k−2
−. . . −2 −1
= 2
k
C(1) +kn −
k−1
¸
j=0
2
j
= 2
k
0 +kn −(2
k
−1) (Eq. 2.10, page 78)
= kn −2
k
+ 1
= nlog n −n + 1.
As a result, we have the following observation:
The Divide and Conquer Paradigm 169
Observation 6.1 The total number of element comparisons performed
by Algorithm mergesort to sort an array of size n, where n is a power of
2, is between (nlog n)/2 and nlog n−n+1. These are exactly the numbers
stated in Observation 1.5 for Algorithm bottomupsort.
As described before, this is no coincidence, as Algorithm bottomup-
sort performs the same element comparisons as Algorithm mergesort
when n is a power of 2. Section 6.6.4 shows empirical results in which the
number of comparisons performed by the two algorithms are close to each
other when n is not a power of 2.
If n is any arbitrary positive integer (not necessarily a power of 2), the
recurrence relation for C(n), the number of element comparisons performed
by Algorithm mergesort, becomes
C(n) =

0 if n = 1
C(n/2|) +C(n/2|) +bn if n ≥ 2
for some nonnegative constant b. By Theorem 2.6, the solution to this
recurrence is C(n) = Θ(nlog n).
Since the operation of element comparison is a basic operation in the
algorithm, it follows that the running time of Algorithm mergesort is
T(n) = Θ(nlog n). By Theorem 12.2, the running time of any algorithm for
sorting by comparisons is Ω(nlog n). It follows that Algorithm mergesort
is optimal.
Clearly, as in Algorithm bottomupsort, the algorithm needs Θ(n) of
space for carrying out the merges. It is not hard to see that the space needed
for the recursive calls is Θ(n) (Exercise 6.12). It follows that the space
complexity of the algorithm is Θ(n). The following theorem summarizes
the main results of this section.
Theorem 6.3 Algorithm mergesort sorts an array of n elements in
time Θ(nlog n) and space Θ(n).
6.4 The Divide and Conquer Paradigm
Now, since we have at our disposal Algorithm bottomupsort, why resort
to a recursive algorithm such as mergesort, especially if we take into
account the amount of extra space needed for the stack, and the extra time
170 Divide and Conquer
brought about by the overhead inherent in handling recursive calls? From
a practical point of view, there does not seem to be any reason to favor
a recursive algorithm on its equivalent iterative version. However, from a
theoretical point of view, recursive algorithms share the merits of being
easy to state, grasp and analyze. To see this, compare the pseudocode of
Algorithm mergesort with that of bottomupsort. It takes much more
time to debug the latter, or even to comprehend the idea behind it. This
suggests that a designer of an algorithm might be better oﬀ starting with
an outline of a recursive description, if possible, which may afterwards be
reﬁned and converted into an iterative algorithm. Note that this is always
possible, as every recursive algorithm can be converted into an iterative
algorithm which functions in exactly the same way on every instance of
the problem. In general, the divide and conquer paradigm consists of the
following steps.
(a) The divide step. In this step of the algorithm, the input is partitioned
into p ≥ 1 parts, each of size strictly less than n, the size of the original
instance. The most common value of p is 2, although other small constants
greater than 2 are not uncommon. We have already seen an example of
the case when p = 2, i.e., Algorithm mergesort. If p = 1 as in Algorithm
binarysearchrec, then part of the input is discarded and the algorithm
recurses on the remaining part. This case is equivalent to saying that the
input data is divided into two parts, where one part is discarded; note that
some work must be done in order to discard some of the elements. p may
also be as high as log n, or even n

, where is some constant, 0 < < 1.
(b) The conquer step. This step consists of performing p recursive call(s)
if the problem size is greater than some predeﬁned threshold n
0
. This
threshold is derived by mathematical analysis of the algorithm. Once it is
found, it can be increased by any constant amount without aﬀecting the
time complexity of the algorithm. In Algorithmmergesort, although n
0
=
1, it can be set to any positive constant without aﬀecting the Θ(nlog n) time
complexity. This is because the time complexity, by deﬁnition, is concerned
with the behavior of the algorithm when n approaches inﬁnity. For example,
we can modify mergesort so that when n ≤ 16, the algorithm uses a
straightforward (iterative) sorting algorithm, e.g. insertionsort. We can
increase it to a much larger value, say 1000. However, after some point,
the behavior of the algorithm starts to degrade. An (approximation to the)
optimal threshold may be found empirically by ﬁne tuning its value until
The Divide and Conquer Paradigm 171
the desired constant is found. It should be emphasized, however, that in
some algorithms, the threshold may not be as low as 1, that is, it must be
greater than some constant that is usually found by a careful analysis of
the algorithm. An example of this is the median ﬁnding algorithm which
will be introduced in Sec. 6.5. It will be shown that the threshold for that
particular algorithm must be relatively high in order to guarantee linear
running time.
(c) The combine step
∗
. In this step, the solutions to the p recursive call(s)
are combined to obtain the desired output. In Algorithm mergesort,
this step consists of merging the two sorted sequences obtained by the two
recursive calls using Algorithm merge. The combine step in a divide-and-
conquer algorithm may consist of merging, sorting, searching, ﬁnding the
maximum or minimum, matrix addition, etc.
The combine step is very crucial to the performance of virtually all
divide-and-conquer algorithms, as the eﬃciency of the algorithm is largely
dependent on how judiciously this step is implemented. To see this, sup-
pose that Algorithm mergesort uses an algorithm that merges two sorted
arrays of size n/2 each in time Θ(nlog n). Then, the recurrence relation
that describes the behavior of this modiﬁed sorting algorithm becomes
T(n) =

0 if n = 1
2C(n/2) +bnlog n if n ≥ 2,
for some nonnegative constant b. By Theorem 2.4, the solution to this
recurrence is T(n) = Θ(nlog
2
n), which is asymptotically larger than the
time complexity of Algorithm mergesort by a factor of log n.
On the other hand, the divide step is invariant in almost all divide-and-
conquer algorithms: Partition the input data into p parts, and proceed to
the conquer step. In many divide-and-conquer algorithms, it takes O(n)
time or even only O(1) time. For example, the time taken by Algorithm
mergesort to divide the input into two halves is constant; it is the time
needed to compute mid. In quicksort algorithm, which will be introduced
in Sec. 6.6, it is the other way around: The divide step requires Θ(n) time,
whereas the combine step is nonexistent.
In general, a divide-and-conquer algorithm has the following format.
∗
Sometimes, this step is referred to as the merge step; this has nothing to do with the
process of merging two sorted sequences as in Algorithm mergesort.
172 Divide and Conquer
(1) If the size of the instance I is “small”, then solve the problem
using a straightforward method and return the answer. Otherwise,
continue to the next step.
(2) Divide the instance I into p subinstances I
1
, I
2
, . . . , I
p
of approxi-
mately the same size.
(3) Recursively call the algorithm on each subinstance I
j
, 1 ≤ j ≤ p,
to obtain p partial solutions.
(4) Combine the results of the p partial solutions to obtain the solution
to the original instance I. Return the solution of instance I.
The overall performance of a divide-and-conquer algorithm is very sen-
sitive to changes in these steps. In the ﬁrst step, the threshold should be
chosen carefully. As discussed before, it may need to be ﬁne tuned until a
reasonable value is found and no more adjustment is needed. In the second
step, the number of partitions should be selected appropriately so as to
achieve the asymptotically minimum running time. Finally, the combine
step should be as eﬃcient as possible.
6.5 Selection: Finding the Median and the kth Smallest
Element
The median of a sequence of n sorted numbers A[1..n] is the “middle”
element. If n is odd, then the middle element is the (n+1)/2th element in
the sequence. If n is even, then there are two middle elements occurring at
positions n/2 and n/2 + 1. In this case, we will choose the n/2th smallest
element. Thus, in both cases, the median is the n/2|th smallest element.
A straightforward method of ﬁnding the median is to sort all elements
and pick the middle one. This takes Ω(nlog n) time, as any comparison-
based sort process must spend at least this much in the worst case (Theo-
rem 12.2).
It turns out that the median, or in general the kth smallest element, in a
set of n elements can be found in optimal linear time. This problem is also
known as the selection problem. The basic idea is as follows. Suppose after
the divide step of every recursive call in a recursive algorithm, we discard
a constant fraction of the elements and recurse on the remaining elements.
Then, the size of the problem decreases geometrically. That is, in each call,
the size of the problem is reduced by a constant factor. For concreteness,
Selection: Finding the Median and the kth Smallest Element 173
let us assume that an algorithm discards 1/3 of whatever objects it is pro-
cessing and recurses on the remaining 2/3. Then, the number of elements
becomes 2n/3 in the second call, 4n/9 in the third call, 8n/27 in the fourth
call, and so on. Now, suppose that in each call, the algorithm does not
spend more than a constant time f ¸or each element. Then, the overall time
spent on processing all elements gives rise to the geometric series
cn + (2/3)cn + (2/3)
2
cn +. . . + (2/3)
j
cn +. . . ,
where c is some appropriately chosen constant. By Eq. 2.12, this quantity
is less than
∞
¸
j=0
cn(2/3)
j
= 3cn = Θ(n).
This is exactly what is done in the selection algorithm. Algorithm se-
lect shown below for ﬁnding the kth smallest element behaves in the same
way. First, if the number of elements is less than 44, a predeﬁned thresh-
old, then the algorithm uses a straightforward method to compute the kth
smallest element. The choice of this threshold will be apparent later when
we analyze the algorithm. The next step partitions the elements into n/5|
groups of ﬁve elements each. If n is not a multiple of 5, the remaining
elements are excluded, and this should not aﬀect the performance of the
algorithm. Each group is sorted and its median, the third element, is ex-
tracted. Next, the median of these medians, denoted by mm, is computed
recursively. Step 6 of the algorithm partitions the elements in A into three
arrays: A
1
, A
2
and A
3
, which, respectively, contain those elements less
than, equal to and greater than mm. Finally, in Step 7, it is determined in
which of the three arrays the kth smallest element occurs, and depending
on the outcome of this test, the algorithm either returns the kth smallest
element, or recurses on either A
1
or A
3
.
Example 6.1 For the sake of this example, let us temporarily change the
threshold in the algorithm from 44 to a smaller number, say 6. Suppose we want
to ﬁnd the median of the following 25 numbers: 8, 33, 17, 51, 57, 49, 35, 11, 25,
37, 14, 3, 2, 13, 52, 12, 6, 29, 32, 54, 5, 16, 22, 23, 7. Let A[1..25] be this sequence
of numbers and k = ]25/2 = 13. We ﬁnd the 13th smallest element in A as
follows.
First, we divide the set of numbers into 5 groups of 5 elements each: (8,
33, 17, 51, 57), (49, 35, 11, 25, 37), (14, 3, 2, 13, 52), (12, 6, 29, 32, 54),
174 Divide and Conquer
Algorithm 6.4 select
Input: An array A[1..n] of n elements and an integer k, 1 ≤ k ≤ n.
Output: The kth smallest element in A.
1. select(A, 1, n, k)
Procedure select(A, low, high, k)
1. p←high −low + 1
2. if p < 44 then sort A and return (A[k])
3. Let q = ]p/5. Divide A into q groups of 5 elements each. If 5 does
not divide p, then discard the remaining elements.
4. Sort each of the q groups individually and extract its median. Let
the set of medians be M.
5. mm←select(M, 1, q, ]q/2) ¦mm is the median of medi-
ans
¦
6. Partition A[low..high] into three arrays:
A
1
= ¦a [ a < mm¦
A
2
= ¦a [ a = mm¦
A
3
= ¦a [ a > mm¦
7. case
[A
1
[ ≥ k: return select(A
1
, 1, [A
1
[, k)
[A
1
[ +[A
2
[ ≥ k: return mm
[A
1
[ +[A
2
[ < k: return select(A
3
, 1, [A
3
[, k −[A
1
[ −[A
2
[)
8. end case
(5, 16, 22, 23, 7). Next, we sort each group in increasing order: (8, 17, 33,
51, 57), (11, 25, 35, 37, 49), (2, 3, 13, 14, 52), (6, 12, 29, 32, 54), (5, 7,
16, 22, 23). Now, we extract the median of each group and form the set of
medians: M = ¦33, 35, 13, 29, 16¦. Next, we use the algorithm recursively to
ﬁnd the median of medians in M: mm = 29. Now, we partition A into three
sequences: A
1
= ¦8, 17, 11, 25, 14, 3, 2, 13, 12, 6, 5, 16, 22, 23, 7¦, A
2
= ¦29¦ and
A
3
= ¦33, 51, 57, 49, 35, 37, 52, 32, 54¦. Since 13 ≤ 15 = [A
1
[, the elements in
A
2
and A
3
are discarded, and the 13th element must be in A
1
. We repeat
the same procedure above, so we set A = A
1
. We divide the elements into
3 groups of 5 elements each: (8,17,11,25,14),(3,2,13,12,6),(5,16,22,23,7). After
sorting each group, we ﬁnd the new set of medians: M = ¦14, 6, 16¦. Thus, the
new median of medians mm is 14. Next, we partition A into three sequences:
A
1
= ¦8, 11, 3, 2, 13, 12, 6, 5, 7¦, A
2
= ¦14¦ and A
3
= ¦17, 25, 16, 22, 23¦. Since
13 > 10 = [A
1
[ +[A
2
[, we set A = A
3
and ﬁnd the 3rd element in A (3 = 13−10).
The algorithm will return A[3] = 22. Thus, the median of the numbers in the
given sequence is 22.
Selection: Finding the Median and the kth Smallest Element 175
6.5.1 Analysis of the selection algorithm
It is not hard to see that Algorithm select correctly computes the kth
smallest element. Now, we analyze the running time of the algorithm.
Consider Fig. 6.2 in which a number of elements have been divided into
5-element groups with the elements in each group ordered from bottom to
top in increasing order.
increasing
order
Set of group medians
sorted in increasing
order from left to right
All elements here are
less than or equal to
the median of medians
All elements here are
greater than or equal to
the median of medians
ii
W
A Y
Z
Fig. 6.2 Analysis of Algorithm select.
Furthermore, these groups have been aligned in such a way that their
medians are in increasing order from left to right. It is clear from the ﬁgure
that all elements enclosed within the rectangle labeled W are less than or
equal to mm, and all elements enclosed within the rectangle labeled X are
greater than or equal to mm. Let A

1
denote the set of elements that are
less than or equal to mm, and A

3
the set of elements that are greater than
or equal to mm. In the algorithm, A
1
is the set of elements that are strictly
less than mm and A
3
is the set of elements that are strictly greater than
mm. Since A

3
[ ≥
3
2
n/5| and [A
1
[ ≤ 0.7n + 1.2.
Thus, we have established upperbounds on the number of elements in A
1
and A
3
, i.e., the number of elements less than mm and the number of el-
ements greater than mm, which cannot exceed roughly 0.7n, a constant
fraction of n.
Now we are in a position to estimate T(n), the running time of the
algorithm. Steps 1 and 2 of procedure select in the algorithm cost Θ(1)
time each. Step 3 costs Θ(n) time. Step 4 costs Θ(n) time, as sorting each
group costs a constant amount of time. In fact sorting each group costs no
more than seven comparisons. The cost of Step 5 is T(n/5|). Step 6 takes
Θ(n) time. By the above analysis, the cost of Step 7 is at most T(0.7n+1.2).
Now we wish to express this ratio in terms of the ﬂoor function and get rid of
the constant 1.2. For this purpose, let us assume that 0.7n+1.2 ≤ 0.75n|.
Then, this inequality will be satisﬁed if 0.7n + 1.2 ≤ 0.75n − 1, that is if
n ≥ 44. This is why we have set the threshold in the algorithm to 44. We
conclude that the cost of this step is at most T(0.75n|) for n ≥ 44. This
analysis implies the following recurrence for the running time of Algorithm
select.
T(n) ≤

c if n < 44
T(n/5|) +T(3n/4|) +cn if n ≥ 44,
for some constant c that is suﬃciently large. Since (1/5) + (3/4) < 1, it
follows by Theorem 2.7, that the solution to this recurrence is T(n) = Θ(n).
In fact, by Example 2.27,
T(n) ≤
cn
1 −1/5 −3/4
= 20cn.
Note that each ratio > 0.7n results in a diﬀerent threshold. For instance,
choosing 0.7n + 1.2 ≤ 0.71n| results in a threshold of about 220. The
following theorem summarizes the main result of this section.
Quicksort 177
Theorem 6.4 The kth smallest element in a set of n elements drawn
from a linearly ordered set can be found in Θ(n) time. In particular, the
median of n elements can be found in Θ(n) time.
It should be emphasized, however, that the multiplicative constant in
the time complexity of the algorithm is too large. In Sec. 14.4, we will
present a simple randomized selection algorithm with Θ(n) expected run-
ning time and a small multiplicative constant. Also, Algorithm select
can be rewritten without the need for the auxiliary arrays A
1
, A
2
and A
3
(Exercise 6.26).
6.6 Quicksort
In this section, we describe a very popular and eﬃcient sorting algo-
rithm: quicksort. This sorting algorithm has an average running time
of Θ(nlog n). One advantage of this algorithm over Algorithm mergesort
is that it sorts the elements in place, i.e., it does not need auxiliary storage
for the elements to be sorted. Before we describe the sorting algorithm, we
need the following partitioning algorithm, which is the basis for Algorithm
quicksort.
6.6.1 A partitioning algorithm
Let A[low..high] be an array of n numbers, and x = A[low]. We consider
the problem of rearranging the elements in A so that all elements less
than or equal to x precede x which in turn precedes all elements greater
than x. After permuting the elements in the array, x will be A[w] for
some w, low ≤ w ≤ high. For example, if A = 5 3 9 2 7 1 8 ,
and low = 1 and high = 7, then after rearranging the elements we will have
A = 1 3 2 5 7 9 8 . Thus, after the elements have been rearranged,
w = 4. The action of rearrangement is also called splitting or partitioning
around x, which is called the pivot or splitting element.
Deﬁnition 6.1 We say that an element A[j] is in its proper position or
correct position if it is neither smaller than the elements in A[low..j − 1]
nor larger than the elements in A[j + 1..high].
An important observation is the following:
178 Divide and Conquer
Observation 6.2 After partitioning an array A using x ∈ A as a pivot,
x will be in its correct position.
In other words, if we sort the elements in A in nondecreasing order after
they have been rearranged, then we will still have A[w] = x. Note that it
is fairly simple to partition a given array A[low..high] if we are allowed
to use another array B[low..high] as an auxiliary storage. What we are
interested in is carrying out the partitioning without an auxiliary array. In
other words, we are interested in rearranging the elements of A in place.
There are several ways to achieve this from which we choose the method
described formally in Algorithm split.
Algorithm 6.5 split
Input: An array of elements A[low..high].
Output: (1)A with its elements rearranged, if necessary, as described above.
(2) w, the new position of the splitting element A[low].
1. i ←low
2. x←A[low]
3. for j ←low + 1 to high
4. if A[j] ≤ x then
5. i ←i + 1
6. if i ,= j then interchange A[i] and A[j]
7. end if
8. end for
9. interchange A[low] and A[i]
10. w←i
11. return A and w
Throughout the execution of the algorithm, we maintain two pointers
i and j that are initially set to low and low + 1, respectively. These two
pointers move from left to right so that after each iteration of the for loop,
we have (see Fig. 6.3(a)):
(1) A[low] = x.
(2) A[k]≤ x for all k, low ≤ k ≤ i.
(3) A[k]> x for all k, i < k ≤ j.
After the algorithm scans all elements, it interchanges the pivot with
A[i], so that all elements smaller than or equal to the pivot are to its left
and all elements larger than the pivot are to its right (see Fig. 6.3(b)).
Finally, the algorithm sets w, the position of the pivot, to i.
Quicksort 179
(b) After the algorithm terminates.
(a) After each iteration of the for loop.
a
j ,
?
|on /jo/
|on /jo/
a
n
a a
a a
Fig. 6.3 The behavior of Algorithm split.
Example 6.2 To aid in the understanding of the algorithm, we apply it to
the input array 5 7 1 6 4 8 3 2 . The working of the algorithm on
this input is illustrated in Fig. 6.4. Figure 6.4.(a) shows the input array. Here
low = 1 and high = 8, and the pivot is x = 5 = A[1]. Initially, i and j point to
elements A[1] and A[2], respectively (see Fig. 6.4.a). To start the partitioning, j
is moved to the right, and since A[3] = 1 ≤ 5 = x, i is incremented and then A[i]
and A[j] are interchanged as shown in Fig. 6.4.(b). Similarly, j is incremented
twice and then A[3] and A[5] are interchanged as shown in Fig. 6.4.(c). Next,
j is moved to the right where an element that is less than x, namely A[7] = 3
is found. Again, i is incremented and A[4] and A[7] are interchanged as shown
in Fig. 6.4.(d). Once more j is incremented and since A[8] = 2 is less than the
pivot, i is incremented and then A[5] and A[8] are interchanged (see Fig. 6.4.e).
Finally, before the procedure ends, the pivot is moved to its proper position by
interchanging A[i] with A[1] as shown in Fig. 6.4.(f).
The following observation is easy to verify:
Observation 6.3 The number of element comparisons performed by Al-
gorithm split is exactly n −1. Thus, its time complexity is Θ(n).
Finally, we note that the only extra space used by the algorithm is that
needed to hold its local variables. Therefore, the space complexity of the
algorithm is Θ(1).
6.6.2 The sorting algorithm
In its simplest form, Algorithm quicksort can be summarized as follows.
The elements A[low..high] to be sorted are rearranged using Algorithm
180 Divide and Conquer
6 7 1 5 4 8 3 2
(c)
,
j
6 7 1 5 4 8 3 2
(d)
,
j
6 7 1 5 4 8 3 2
(e)
,
j
6 7 1 5 4 8 3 2
(f)
,
j
6 7 1 5 4 8 3 2
(b)
,
j
6 7 1 5 4 8 3 2
(a)
,
j
Fig. 6.4 Example of partitioning a sequence of numbers using Algorithm split.
split so that the pivot element, which is always A[low], occupies its correct
position A[w], and all elements that are less than or equal to A[w] occupy
the positions A[low..w − 1], while all elements that are greater than A[w]
occupy the positions A[w + 1..high]. The subarrays A[low..w − 1] and
A[w+1..high] are then recursively sorted to produce the entire sorted array.
The formal algorithm is shown as Algorithm quicksort.
Algorithm 6.6 quicksort
Input: An array A[1..n] of n elements.
Output: The elements in A sorted in nondecreasing order.
1. quicksort(A, 1, n)
Procedure quicksort(A, low, high)
1. if low < high then
2. split(A[low..high], w) ¦w is the new position of A[low]¦
3. quicksort(A, low, w −1)
4. quicksort(A, w + 1, high)
5. end if
The relationship of Algorithm split to Algorithm quicksort is similar
to the relationship of Algorithm merge to Algorithm mergesort; both
Quicksort 181
sorting algorithms consist of a series of calls to one of these two basic al-
gorithms, namely merge and split. However, there is a subtle diﬀerence
between the two from the algorithmic point of view: In Algorithm merge-
sort, merging the sorted sequences belongs to the combine step, whereas
splitting in Algorithm quicksort belongs to the divide step. Indeed, the
combine step in Algorithm quicksort is nonexistent.
Example 6.3 Suppose we want to sort the array 4 6 3 1 8 7 2 5 .
The sequence of splitting the array and its subarrays is illustrated in Fig. 6.5.
Each pair of arrays in the ﬁgure corresponds to an input to and output of Algo-
rithm split. Darkened boxes are used for the pivots. For example, in the ﬁrst
call, Algorithm split was presented with the above 8-element array. By Obser-
vation 6.2, after splitting the array, 4 will occupy its proper position, namely
position 4. Consequently, the problem now reduces to sorting the two subarrays
2 3 1 and 8 7 6 5 . Since calls are implemented in a preorder fashion,
the second call induces the third call on input 1 . Another call on only one ele-
ment, namely 3 is executed. At this point, the ﬂow of control backs up to the
ﬁrst call and another call on the subarray 8 7 6 5 is initiated. Continuing
this way, the array is ﬁnally sorted after eight calls to procedure quicksort.
6.6.3 Analysis of the quicksort algorithm
In this section, we analyze the running time of Algorithm quicksort. We
will show that although it exhibits a running time of Θ(n
2
) in the worst
case, its average time complexity is indeed Θ(nlog n). This together with
the fact that it sorts in place makes it very popular and comparable to
heapsort in practice. Although no auxiliary storage is required by the
algorithm to store the array elements, the space required by the algorithm
is O(n). This is because in every recursive call, the left and right indices of
the right part of the array to be sorted next, namely w+1 and high, must
be stored. It will be left as an exercise to show that the work space needed
by the algorithm varies between Θ(log n) and Θ(n) (Exercise 6.37).
6.6.3.1 The worst case behavior
To ﬁnd the running time of Algorithm quicksort in the worst case, we
only need to ﬁnd one situation in which the algorithm exhibits the longest
running time for each value of n. Suppose that in every call to Algorithm
182 Divide and Conquer
1st call
2nd call
3rd call
4th call
6th call
5th call
7th call
8th call
4 6 3 1 8 7 2 5
2 3 1 4 8 7 6 5
2 3 1
3 1 2
1
1
1 2 3 4 5 6 7 8
8 7 6 5
5 7 6 8
6
6
3
3
5 7 6
5 7 6
7 6
6 7
Fig. 6.5 Example of the execution of Algorithm quicksort.
quicksort, it happens that the pivot, which is A[low], is the smallest
number in the array. This means that Algorithm split will return w = low
and, consequently, there will be only one nontrivial recursive call, the other
call being a call on an empty array. Thus, if Algorithm quicksort is
initiated by the call quicksort(A, 1, n), the next two recursive calls will
be quicksort(A, 1, 0) and quicksort(A, 2, n), with the ﬁrst being a trivial
call. It follows that the worst case happens if the input array is already
sorted in nondecreasing order! In this case the smallest element will always
be chosen as the pivot, and as a result, the following n calls to procedure
quicksort will take place:
quicksort(A, 1, n), quicksort(A, 2, n), . . . , quicksort(A, n, n).
These, in turn, initiate the following nontrivial calls to Algorithm split.
split(A[1..n], w), split(A[2..n], w), . . . split(A[n..n], w).
Quicksort 183
Since the number of comparisons done by the splitting algorithm on
input of size j is j − 1 (Observation 6.3), it follows that the total number
of comparisons performed by Algorithm quicksort in the worst case is
(n −1) + (n −2) +. . . + 1 + 0 =
n(n −1)
2
= Θ(n
2
).
It should be emphasized, however, that this extreme case is not the only
case that leads to a quadratic running time. If, for instance, the algorithm
always selects one of the smallest (or largest) k elements, for any constant
k that is suﬃciently small relative to n, then the algorithm’s running time
is also quadratic.
The worst case running time can be improved to Θ(nlog n) by always
selecting the median as the pivot in linear time as shown in Sec. 6.5. This
is because the splitting of elements is highly balanced; in this case the
two recursive calls have approximately the same number of elements. This
results in the following recurrence for counting the number of comparisons:
C(n) =

0 if n = 1
2C(n/2) + Θ(n) if n > 1,
whose solution is C(n) = Θ(nlog n). However, the hidden constant in
the time complexity of the median ﬁnding algorithm is too high to be used
in conjunction with Algorithm quicksort. Thus, we have the following
theorem:
Theorem 6.5 The running time of Algorithm quicksort is Θ(n
2
) in
the worst case. If, however, the median is always chosen as the pivot, then
its time complexity is Θ(nlog n).
It turns out, however, that Algorithm quicksort as originally stated,
is a fast sorting algorithm in practice (given that the elements to be sorted
are in random order); this is supported by its average time analysis, which
is discussed below. If the elements to be sorted are not in random order,
then choosing the pivot randomly, instead of always using A[low], results
in a very eﬃcient algorithm. This version of Algorithm quicksort will be
presented in Sec. 14.3.
184 Divide and Conquer
6.6.3.2 The average case behavior
It is important to note that the above extreme cases are practically rare,
and in practice the running time of Algorithm quicksort is fast. This
motivates the investigation of its performance on the average. It turns out
that, on the average, its time complexity is Θ(nlog n), and not only that,
but also the multiplicative constant is fairly small. For simplicity, we will
assume that the input elements are distinct. Note that the behavior of
the algorithm is independent of the input values; what matters is their
relative order. For this reason, we may assume without loss of generality,
that the elements to be sorted are the ﬁrst n positive integers 1, 2, . . . , n.
When analyzing the average behavior of an algorithm, it is important to
assume some probability distribution on the input. In order to simplify the
analysis further, we will assume that each permutation of the elements is
equally likely. That is, we will assume that each of the n! permutations of
the numbers 1, 2, . . . , n is equally likely. This ensures that each number in
the array is equally likely to be the ﬁrst element and thus chosen as the
pivot, i.e., the probability that any element of A will be picked as the pivot
is 1/n. Let C(n) denote the number of comparisons done by the algorithm
on the average on an input of size n. From the assumptions stated (all
elements are distinct and have the same probability of being picked as the
pivot), the average cost is computed as follows. By Observation 6.3, Step 2
costs exactly n−1 comparisons. Steps 3 and 4 cost C(w−1) and C(n−w)
comparisons, respectively. Hence, the total number of comparisons is
C(n) = (n −1) +
1
n
n
¸
w=1
(C(w −1) +C(n −w)). (6.1)
Since
n
¸
w=1
C(n −w) = C(n −1) +C(n −2) +. . . +C(0) =
n
¸
w=1
C(w −1),
Eq. 6.1 can be simpliﬁed to
C(n) = (n −1) +
2
n
n
¸
w=1
C(w −1). (6.2)
This recurrence seems to be complicated when compared with the recur-
rences we are used to, as the value of C(n) depends on all its history:
Quicksort 185
C(n −1), C(n −2), . . . , C(0). However, we can remove this dependence as
follows. First, we multiply Eq. 6.2 by n:
nC(n) = n(n −1) + 2
n
¸
w=1
C(w −1). (6.3)
If we replace n by n −1 in Eq. 6.3, we obtain
(n −1)C(n −1) = (n −1)(n −2) + 2
n−1
¸
w=1
C(w −1). (6.4)
Subtracting Eq. 6.4 from Eq. 6.3 and rearranging terms yields
C(n)
n + 1
=
C(n −1)
n
+
2(n −1)
n(n + 1)
. (6.5)
Now, we change to a new variable D, by letting
D(n) =
C(n)
n + 1
.
In terms of the new variable D, Eq. 6.5 can be rewritten as
D(n) = D(n −1) +
2(n −1)
n(n + 1)
, D(1) = 0. (6.6)
Clearly, the solution of Eq. 6.6 is
D(n) = 2
n
¸
j=1
j −1
j(j + 1)
.
We simplify this expression as follows.
2
n
¸
j=1
j −1
j(j + 1)
= 2
n
¸
j=1
2
(j + 1)
−2
n
¸
j=1
1
j
= 4
n+1
¸
j=2
1
j
−2
n
¸
j=1
1
j
= 2
n
¸
j=1
1
j
−
4n
n + 1
= 2 lnn −Θ(1) (Eq. 2.16)
=
2
log e
log n −Θ(1)
186 Divide and Conquer
≈ 1.44log n.
Consequently,
C(n) = (n + 1)D(n) ≈ 1.44nlog n.
We have, in eﬀect, proven the following theorem:
Theorem 6.6 The average number of comparisons performed by Algo-
rithm quicksort to sort an array of n elements is Θ(nlog n).
6.6.4 Comparison of sorting algorithms
Table 6.1 gives the output of a sorting experiment for the average number
of comparisons of ﬁve sorting algorithms using values of n between 500 and
5000.
n selectionsort insertionsort bottomupsort mergesort quicksort
500 124750 62747 3852 3852 6291
1000 499500 261260 8682 8704 15693
1500 1124250 566627 14085 13984 28172
2000 1999000 1000488 19393 19426 34020
2500 3123750 1564522 25951 25111 52513
3000 4498500 2251112 31241 30930 55397
3500 6123250 3088971 37102 36762 67131
4000 7998000 4042842 42882 42859 79432
4500 10122750 5103513 51615 49071 98635
5000 12497500 6180358 56888 55280 106178
Table 6.1 Comparison of sorting algorithms.
The numbers under each sorting algorithm are the counts of the number
of comparisons performed by the respective algorithm. From the table, we
can see that the average number of comparisons performed by Algorithm
quicksort is almost double that of mergesort and bottomupsort.
Multiplication of Large Integers 187
6.7 Multiplication of Large Integers
We have assumed in the beginning that multiplication of integers whose
size is ﬁxed costs a unit amount of time. This is no longer valid when
multiplying two integers of of arbitrary length. As explained in Sec. 1.14,
the input to an algorithm dealing with numbers of variable size is usually
measured in the number of bits or, equivalently, digits. Let u and v be
two n-bit integers. The traditional multiplication algorithm requires Θ(n
2
)
digit multiplications to compute the product of u and v. Using the divide
and conquer technique, this bound can be reduced signiﬁcantly as follows.
For simplicity, assume that n is a power of 2.
Each integer is divided into two parts of n/2 bits each. Then, u and v
can be rewritten as u = w2
n/2
+x and v = y2
n/2
+z (see Fig. 6.6).
u = w2
n/2
+x :
w x
v = y2
n/2
+z :
y z
Fig. 6.6 Multiplication of two large integers.
The product of u and v can be computed as
uv = (w2
n/2
+x)(y2
n/2
+z) = wy2
n
+ (wz +xy)2
n/2
+xz. (6.7)
Note that multiplying by 2
n
amounts to simply shifting by n bits to the left,
which takes Θ(n) time. Thus, in this formula, there are four multiplications
and three additions. This implies the following recurrence:
T(n) =

d if n = 1
3T(n/2) +bn if n > 1
for some appropriately chosen constants b and d > 0. Again, by Theo-
rem 2.5,
T(n) = Θ(n
log 3
) = O(n
1.59
),
a remarkable improvement on the traditional method.
6.8 Matrix Multiplication
Let A and B be two n n matrices. We wish to compute their product
C = AB. In this section, we show how to apply the divide and conquer
strategy to this problem to obtain an eﬃcient algorithm.
6.8.1 The traditional algorithm
In the traditional method, C is computed using the formula
C(i, j) =
n
¸
k=1
A(i, k)B(k, j).
It can be easily shown that this algorithm requires n
3
multiplications and
n
3
−n
2
additions (Exercise 6.46). This results in a time complexity of Θ(n
3
).
6.8.2 Recursive version
Assume that n = 2
k
, k ≥ 0. If n ≥ 2, then A, B and C can be partitioned
into four matrices of dimensions n/2 n/2 each:
A =

A
11
A
12
A
21
A
22

, B =

B
11
B
12
B
21
B
22

, C =

C
11
C
12
C
21
C
22

.
Matrix Multiplication 189
The divide-and-conquer version consists of computing C as deﬁned by the
equation
C =

.
This requires 8 multiplications and 4 additions of n/2 n/2 matrices.
The cost of multiplying two n n matrices is thus 8 times the cost of
multiplying two n/2 n/2 matrices plus 4 times the cost of adding two
n/2 n/2 matrices. In order to count the number of scalar operations, let
a and m denote the costs of scalar addition and multiplication, respectively.
If n = 1, the total cost is just m since we have only one scalar multiplication.
Thus, the total cost of multiplying two n n matrices is governed by the
recurrence
T(n) =

m if n = 1
8T(n/2) + 4(n/2)
2
a if n ≥ 2,
which can be rewritten as
T(n) =

m if n = 1
8T(n/2) +an
2
if n ≥ 2.
By Lemma 2.1,
T(n) =

m+
a2
2
8 −2
2

n
3
−

a2
2
8 −2
2

n
2
= mn
3
+an
3
−an
2
.
Thus, we have, as a result, the following observation:
Observation 6.4 The recursive version requires n
3
multiplications and
n
3
−n
2
additions. These are exactly the ﬁgures stated above for the tradi-
tional method.
It follows that this divide-and-conquer version does not result in an eﬃ-
cient algorithm. On the contrary, it costs more than the traditional method.
The added cost in terms of both time and space comes from the overhead
brought about by recursion. This is reminiscent of the relationship between
algorithms bottomupsort and mergesort stated in Observation 6.1. In
other words, this is simply a recursive version of the traditional method.
The two algorithms diﬀer only in the order in which the matrix elements
are multiplied, otherwise they are identical.
190 Divide and Conquer
6.8.3 Strassen’s algorithm
This algorithm has a o(n
3
) time complexity, i.e., its running time is asymp-
totically less than n
3
. This is a remarkable improvement on the traditional
algorithm. The idea behind this algorithm consists in reducing the num-
ber of multiplications at the expense of increasing the number of additions
and subtractions. In short, this algorithm uses 7 multiplications and 18
additions of n/2 n/2 matrices.
Let
A =

.
Since commutativity of scalar products is not used here, the above for-
mula holds for matrices as well.
Time Complexity
The number of additions used is 18 and the number of multiplications is 7.
This gives rise to the following recurrence for the running time.
T(n) =

(x
1
−x
2
)
2
+ (y
1
−y
2
)
2
is minimum among all pairs of points in S. Here d(p
1
, p
2
) is referred to as
the Euclidean distance between p
1
and p
2
. The brute-force algorithm sim-
ply examines all the possible n(n−1)/2 distances and returns that pair with
smallest separation. In this section, we describe a Θ(nlog n) time algorithm
to solve the closest pair problem using the divide and conquer design tech-
nique. Instead of ﬁnding that pair which realizes the minimum distance,
the algorithm to be developed will only return the distance between them.
Modifying the algorithm to return that pair as well is easy.
The general outline of the algorithm can be summarized as follows.
The ﬁrst step in the algorithm is to sort the points in S by increasing x-
coordinate. Next, the point set S is divided about a vertical line L into two
subsets S
l
and S
r
such that [S
l
[ = [S[/2| and [S
r
[ = [S[/2|. Let L be
the vertical line passing by the x-coordinate of S[n/2|]. Thus, all points
in S
l
are on or to the left of L, and all points in S
r
are on or to the right of
L. Now, recursively, the minimum separations δ
l
and δ
r
of the two subsets
S
l
and S
r
, respectively, are computed. For the combine step, the smallest
separation δ

between a point in S
l
and a point in S
r
is also computed.
Finally, the desired solution is the minimum of δ
l
, δ
r
and δ

.
As in most divide-and-conquer algorithms, most of the work comes from
the combine step. At this point, it is not obvious how to implement this
step. The crux of this step is in computing δ

. The na¨ıve method which
computes the distance between each point in S
l
and each point in S
r
re-
quires Ω(n
2
), and hence an eﬃcient approach to implement this step must
be found.
Let δ = min¦δ
l
, δ
r
¦. If the closest pair consists of some point p
l
in S
l
and some point p
r
in S
r
, then p
l
and p
r
must be within distance δ of the
dividing line L. Thus, if we let S

l
and S

r
denote, respectively, the points
in S
l
and in S
r
within distance δ of L, then p
l
must be in S

r
= S
r
. The crucial
observation is that not all these O(n
2
) comparisons are indeed necessary; we
only need to compare each point p in S
l
, say, with those within distance δ. A
close inspection of Fig. 6.7 reveals that the points lying within the two strips
of width δ around L have a special structure. Suppose that δ

≤ δ. Then
there exist two points p
l
∈ S

l
and p
r
∈ S

r
such that d(p
l
, p
r
) = δ

. It follows
that the vertical distance between p
l
and p
r
is at most δ. Furthermore, since
p
l
∈ S

l
and p
r
∈ S

r
, these two points are inside or on the boundary of a
δ 2δ rectangle centered around the vertical line L (see Fig. 6.8).
1
T
Fig. 6.8 Further illustration of the combine step.
Let T be the set of points within the two vertical strips. Referring again
194 Divide and Conquer
to Fig. 6.8, if the distance between any two points in the δ 2δ rectangle
must be at most δ, then the rectangle can accommodate at most eight
points: at most four points from S
l
and at most four points from S
r
. The
maximum number is attained when one point from S
l
coincides with one
point from S
r
at the intersection of L with the top of the rectangle, and
one point from S
l
coincides with one point from S
r
at the intersection of
L with the bottom of the rectangle. This implies the following important
observation:
Observation 6.5 Each point in T needs to be compared with at most
seven points in T.
The above observation gives only an upper bound on the number of
points to be compared with each point p in T, but does not give any in-
formation as to which points are to be compared with p. A moment of
reﬂection shows that p must be compared with its neighbors in T. To
ﬁnd such neighbors, we resort to sorting the points in T by increasing y-
coordinate. After that, it is not hard to see that we only need to compare
each point p in T with those seven points following p in increasing order of
their y-coordinates.
6.9.1 Time complexity
Let us analyze the running time of the algorithm developed so far. Sorting
the points in S requires O(nlog n) time. Dividing the points into S
l
and
S
r
takes Θ(1) time, as the points are sorted. As to the combine step, we
see that it consists of sorting the points in T and comparing each point
with at most seven other points. Sorting costs [T[ log [T[ = O(nlog n), and
there are at most 7n comparisons. Thus, the combine step takes Θ(nlog n)
time in the worst case. Noting that if the number of points is three, then
only three comparisons are needed to compute the minimum separation
between them, the recurrence relation for the performance of the algorithm
becomes
T(n) =

1 if n = 2
3 if n = 3
2T(n/2) +O(nlog n) if n > 3.
By Theorem 2.4, the solution to the above recurrence is T(n) =
O(nlog
2
n), which is not the desired bound. We observe that if we re-
Exercises 195
duce the time taken by the combine step to Θ(n), then the time complexity
of the algorithm will be Θ(nlog n). This can be achieved by a process
called presorting, i.e., the elements in S are sorted by their y-coordinates
once and for all and stored in an array Y . Each time we need to sort T in
the combine step, we only need to extract its elements from Y in Θ(n) time.
This is easy to do since the points in T are those points in Y within distance
δ from the vertical line L. This modiﬁcation reduces the time required by
the combine step to Θ(n). Thus, the recurrence relation becomes
T(n) =

1 if n = 2
3 if n = 3
2T(n/2) + Θ(n) if n > 3.
The solution to this familiar recurrence is the desired Θ(nlog n) bound.
The above discussion implies Algorithm closestpair. In the algorithm,
for a point p, x(p) denotes the x-coordinate of point p.
The following theorem summarizes the main result. Its proof is em-
bedded in the description of the algorithm and the analysis of its running
time.
Theorem 6.7 Given a set S of n points in the plane, Algorithm clos-
estpair ﬁnds a pair of points in S with minimum separation in Θ(nlog n)
time.
6.10 Exercises
6.1. Modify Algorithm minmax so that it works when n is not a power of 2. Is
the number of comparisons performed by the new algorithm ]3n/2 −2
even if n is not a power of 2? Prove your answer.
6.2. Consider Algorithm slowminmax which is obtained from Algorithm
minmax by replacing the test
if high −low = 1
by the test
if high = low
and making some other changes in the algorithm accordingly. Thus, in
Algorithm slowminmax, the recursion is halted when the size of the
input array is 1. Count the number of comparisons required by this al-
gorithm to ﬁnd the minimum and maximum of an array A[1..n], where
196 Divide and Conquer
Algorithm 6.7 closestpair
Input: A set S of n points in the plane.
Output: The minimum separation realized by two points in S.
1. Sort The points in S in nondecreasing order of their x-coordinates.
2. Y ←The points in S sorted in nondecreasing order of their y-
coordinates.
3. δ ←cp(1, n)
Procedure cp(low, high)
1. if high −low +1 ≤ 3 then compute δ by a straightforward method.
2. else
3. mid ←](low + high)/2
4. x
0
←x(S[mid])
5. δ
l
←cp(low, mid)
6. δ
r
←cp(mid + 1, high)
7. δ ← min¦δ
l
, δ
r
¦
8. k←0
9. for i ←1 to n ¦Extract T from Y ¦
10. if [x(Y [i]) −x
0
[ ≤ δ then
11. k←k + 1
12. T[k] ←Y [i]
13. end if
14. end for ¦k is the size of T¦
15. δ

←2δ ¦Initialize δ

to any number greater than δ¦
16. for i ←1 to k −1 ¦Compute δ

¦
17. for j ←i + 1 to min¦i + 7, k¦
18. if d(T[i], T[j]) < δ

then δ

←d(T[i], T[j])
19. end for
20. end for
21. δ ← min¦δ, δ

¦
22. end if
23. return δ
n is a power of 2. Explain why the number of comparisons in this al-
gorithm is greater than that in Algorithm minmax. (Hint: In this case,
the initial condition is C(1) = 0).
6.3. Derive an iterative minimax algorithm that ﬁnds both the minimum and
maximum in a set of n elements using only 3n/2 −2 comparisons, where
n is a power of 2.
6.4. Give a divide-and-conquer version of Algorithm linearsearch given in
Sec. 1.3. The algorithm should start by dividing the input elements into
approximately two halves. How much work space is required by the
algorithm?
Exercises 197
6.5. Give a divide-and-conquer algorithm to ﬁnd the sum of all numbers in
an array A[1..n] of integers. The algorithm should start by dividing the
input elements into approximately two halves. How much work space is
required by the algorithm?
6.6. Let A[1..n] be an array of n integers and x an integer. Derive a divide-
and-conquer algorithm to ﬁnd the frequency of x in A, i.e., the number
of times x appears in A. What is the time complexity of your algorithm?
6.7. Modify Algorithm binarysearchrec so that it searches for two keys.
In other words, given an array A[1..n] of n elements and two elements x
1
and x
2
, the algorithm should return two integers k
1
and k
2
representing
the positions of x
1
and x
2
, respectively, in A.
6.8. Design a search algorithm that divides a sorted array into one third
and two thirds instead of two halves as in Algorithm binarysearchrec.
Analyze the time complexity of the algorithm.
6.9. Modify Algorithm binarysearchrec so that it divides the sorted array
into three equal parts instead of two as in Algorithm binarysearchrec.
In each iteration, the algorithm should test the element x to be searched
for against two entries in the array. Analyze the time complexity of the
algorithm.
6.10. Use Algorithm mergesort to sort the array
(a) 32 15 14 15 11 17 25 51 .
(b) 12 25 17 19 51 32 45 18 22 37 15 .
6.11. Use mathematical induction to prove the correctness of Algorithm
mergesort. Assume that Algorithm merge works correctly.
6.12. Show that the space complexity of Algorithm mergesort is Θ(n).
6.13. It was shown in Sec. 6.3 that algorithms bottomupsort and mergesort
are very similar. Give an example of an array of numbers in which
(a) Algorithm bottomupsort and Algorithm mergesort perform the
same number of element comparisons.
(b) Algorithm bottomupsort performs more element comparisons
than Algorithm mergesort.
(c) Algorithm bottomupsort performs fewer element comparisons
than Algorithm mergesort.
6.14. Consider the following modiﬁcation of Algorithm mergesort. The algo-
rithm ﬁrst divides the input array A[low..high] into four parts A
1
, A
2
, A
3
and A
4
instead of two. It then sorts each part recursively, and ﬁnally
merges the four sorted parts to obtain the original array in sorted order.
Assume for simplicity that n is a power of 4.
(a) Write out the modiﬁed algorithm.
198 Divide and Conquer
(b) Analyze its running time.
6.15. What will be the running time of the modiﬁed algorithm in Exercise 6.14
if the input array is divided into k parts instead of 4? Here, k is a ﬁxed
positive integer greater than 1.
6.16. Consider the following modiﬁcation to Algorithm mergesort. We apply
the algorithm on the input array A[1..n] and continue the recursive calls
until the size of a subinstance becomes relatively small, say m or less.
At this point, we switch to Algorithm insertionsort and apply it on
the small instance. So, the ﬁrst test of the modiﬁed algorithm will look
like the following:
if high −low + 1 ≤ m then insertionsort(A[low..high]).
What is the largest value of m in terms of n such that the running time
of the modiﬁed algorithm will still be Θ(nlog n)? You may assume for
simplicity that n is a power of 2.
6.17. Use Algorithm select to ﬁnd the kth smallest element in the list of
numbers given in Example 6.1, where
(a) k = 1. (b) k = 9. (c) k = 17. (d) k = 22. (e) k = 25.
6.18. What will happen if in Algorithm select the true median of the elements
is chosen as the pivot instead of the median of medians? Explain.
6.19. Let A[1..105] be a sorted array of 105 integers. Suppose we run Algorithm
select to ﬁnd the 17th element in A. How many recursive calls to
Procedure select will there be? Explain your answer clearly.
6.20. Explain the behavior of Algorithm select if the input array is already
sorted in nondecreasing order. Compare that to the behavior of Algo-
rithm binarysearchrec.
6.21. In Algorithm select, groups of size 5 are sorted in each invocation of
the algorithm. This means that ﬁnding a procedure that sorts a group
of size 5 that uses the fewest number of comparisons is important. Show
that it is possible to sort ﬁve elements using only seven comparisons.
6.22. One reason that Algorithm select is ineﬃcient is that it does not make
full use of the comparisons that it makes: After it discards one portion
of the elements, it starts on the subproblem from scratch. Give a pre-
cise count of the number of comparisons the algorithm performs when
presented with n elements. Note that it is possible to sort ﬁve elements
using only seven comparisons (see Exercise 6.21).
6.23. Based on the number of comparisons counted in Exercise 6.22, determine
for what values of n one should use a straightforward sorting method and
extract the kth element directly.
6.24. Let g denote the size of each group in Algorithm select for some positive
integer g ≥ 3. Derive the running time of the algorithm in terms of g.
Exercises 199
What happens when g is too large compared to the value used in the
algorithm, namely 5?
6.25. Which of the following group sizes 3, 4, 5, 7, 9, 11 guarantees Θ(n)
worst case performance for Algorithm select? Prove your answer. (See
Exercise 6.24).
6.26. Rewrite Algorithm select using Algorithm split to partition the input
array. Assume for simplicity that all input elements are distinct. What
is the advantage of the modiﬁed algorithm?
6.27. Let A[1..n] and B[1..n] be two arrays of distinct integers sorted in in-
creasing order. Give an eﬃcient algorithm to ﬁnd the median of the 2n
elements in both A and B. What is the running time of your algorithm?
6.28. Make use of the algorithm obtained in Exercise 6.27 to device a divide-
and-conquer algorithm for ﬁnding the median in an array A[1..n]. What
is the time complexity of your algorithm? (Hint: Make use of Algorithm
mergesort).
6.29. Consider the problem of ﬁnding all the ﬁrst k smallest elements in an
array A[1..n] of n distinct elements. Here, k is not constant, i.e., it is part
of the input. We can solve this problem easily by sorting the elements
and returning A[1..k]. This, however, costs O(nlog n) time. Give a Θ(n)
time algorithm for this problem. Note that running Algorithm select
k times costs Θ(kn) = O(n
2
) time, as k is not constant.
6.30. Consider the multiselection problem: Given a set S of n elements and
a set K of r ranks k
1
, k
2
, . . . , k
r
, ﬁnd the k
1
th, k
2
th, . . . , k
r
th smallest
elements. For example, if K = ¦2, 7, 9, 50¦, the problem is to ﬁnd the
2nd, 7th, 9th and 50th smallest elements. This problem can be solved
trivially in Θ(rn) time by running Algorithm select r times, once for
each rank k
j
, 1 ≤ j ≤ r. Give an O(nlog r) time algorithm to solve this
problem.
6.31. Apply Algorithm split on the array 27 13 31 18 45 16 17 53 .
6.32. Let f(n) be the number of element interchanges that Algorithm split
makes when presented with the input array A[1..n] excluding interchang-
ing A[low] with A[i].
(a) For what input arrays A[1..n] is f(n) = 0?
(b) What is the maximum value of f(n)? Explain when this maximum
is achieved?
6.33. Modify Algorithm split so that it partitions the elements in A[low..high]
around x, where x is the median of ¦A[low], A[](low + high)/2],
A[high]¦. Will this improve the running time of Algorithm quicksort?
Explain.
6.34. Algorithm split is used to partition an array A[low..high] around A[low].
200 Divide and Conquer
Another algorithm to achieve the same result works as follows. The
algorithm has two pointers i and j. Initially, i = low and j = high. Let
the pivot be x = A[low]. The pointers i and j move from left to right
and from right to left, respectively, until it is found that A[i] > x and
A[j] ≤ x. At this point A[i] and A[j] are interchanged. This process
continues until i ≥ j. Write out the complete algorithm. What is the
number of comparisons performed by the algorithm?
6.35. Let A[1..n] be a set of integers. Give an algorithm to reorder the ele-
ments in A so that all negative integers are positioned to the left of all
nonnegative integers. Your algorithm should run in time Θ(n).
6.36. Use Algorithm quicksort to sort the array
(a) 24 33 24 45 12 12 24 12 .
(b) 3 4 5 6 7 .
(c) 23 32 27 18 45 11 63 12 19 16 25 52 14 .
6.37. Show that the work space needed by Algorithm quicksort varies be-
tween Θ(log n) and Θ(n). What is its average space complexity?
6.38. Explain the behavior of Algorithm quicksort when the input is already
sorted in decreasing order. You may assume that the input elements are
all distinct.
6.39. Explain the behavior of Algorithm quicksort when the input array
A[1..n] consists of n identical elements.
6.40. Modify Algorithm quicksort slightly so that it will solve the selection
problem. What is the time complexity of the new algorithm in the worst
case and on the average?
6.41. Give an iterative version of Algorithm quicksort.
6.42. Which of the following sorting algorithms are stable (see Exercise 5.14)?
(a)heapsort (b)mergesort (c)quicksort.
6.43. A sorting algorithm is called adaptive if its running time depends not
only on the number of elements n, but also on their order. Which of the
following sorting algorithms are adaptive?
(a)selectionsort (b)insertionsort (c)bubblesort (d)heapsort
(e)bottomupsort (f)mergesort (g)quicksort (h)radixsort.
6.44. Let x = a + bi and y = c + di be two complex numbers. The product
xy can easily be calculated using four multiplications, that is, xy =
(ac − bd) + (ad + bc)i. Devise a method for computing the product xy
using only three multiplications.
6.45. Write out an algorithm for the traditional algorithm for matrix multipli-
cation described in Sec. 6.8.
Exercises 201
6.46. Show that the traditional algorithm for matrix multiplication described
in Sec. 6.8 requires n
3
multiplications and n
3
− n
2
additions (see Exer-
cise 6.45).
6.47. Explain how to modify Strassen’s algorithm for matrix multiplication so
that it can also be used with matrices whose size is not necessarily a
power of 2.
6.48. Suppose we modify the algorithm for the closest pair problem so that not
each point in T is compared with seven points in T. Instead, every point
to the left of the vertical line L is compared with a number of points to
its right.
(a) What are the necessary modiﬁcations to the algorithm?
(b) How many points to the right of L have to be compared with every
point to its left? Explain.
6.49. Rewrite the algorithm for the closest pair problem without the presorting
step. The time complexity of your algorithm should be Θ(nlog n). (Hint:
Make use of Algorithm mergesort).
6.50. Design a divide-and-conquer algorithm to determine whether two given
binary trees T
1
and T
2
are identical.
6.51. Design a divide-and-conquer algorithm that computes the height of a
binary tree.
6.52. Give a divide-and-conquer algorithm to ﬁnd the second largest element
in an array of n numbers. Derive the time complexity of your algorithm.
6.53. Consider the following algorithm that attempts to ﬁnd a minimum cost
spanning tree MST(G) for a weighted undirected graph G = (V, E) (see
Sec. 8.3). Divide G into two subgraphs G
1
and G
2
of approximately the
same number of vertices. Compute T
1
= MST(G
1
) and T
2
= MST(G
2
).
Find an edge e of minimum weight that connects G
1
with G
2
. Return
T
1
∪ T
2
∪ ¦e¦. Show that this algorithm does not always compute a
spanning tree of minimum weight. What is the shape of the spanning
tree computed by the algorithm?
6.54. Let B be an n n chessboard, where n is a power of 2. Use a divide-
and-conquer argument to describe (in words) how to cover all squares
of B except one with L-shaped tiles. For example, if n = 2, then there
are four squares three of which can be covered by one L-shaped tile, and
if n = 4, then there are 16 squares of which 15 can be covered by 5
L-shaped tiles.
6.55. Use a combinatorial argument to show that if n is a power of 2, then
n
2
≡ 1 (mod 3). (Hint: Use the result of Exercise 6.54).
202 Divide and Conquer
6.11 Bibliographic notes
Algorithm mergesort and quicksort are discussed in detail in
Knuth(1973). Algorithm quicksort is due to Hoare(1962). The linear
time algorithm for selection is due to Blum, Floyd, Pratt, Rivest and Tar-
jan (1972). The algorithm for integer multiplication is due to Karatsuba
and Ofman (1962). Strassen’s algorithm for matrix multiplication is due to
Strassen (1969). While Strassen’s algorithm uses a fast method to multiply
2 2 matrices as the base case, similar algorithms have since been devel-
oped that use more complex base cases. For example, Pan (1978) proposed
a method based on an eﬃcient scheme for multiplying 70 70 matrices.
The exponent in the time complexity has been steadily reduced. This is
only of theoretical interest, as Strassen’s algorithm remains the only one
of practical interest. The algorithm for the closest pair problem is due to
Shamos and can be found in virtually any book on computational geometry.
Chapter 7
Dynamic Programming
7.1 Introduction
In this chapter we study a powerful algorithm design technique that is
widely used to solve combinatorial optimization problems. An algorithm
that employs this technique is not recursive by itself, but the underlying
solution of the problem is usually stated in the form of a recursive function.
Unlike the case in divide-and-conquer algorithms, immediate implementa-
tion of the recurrence results in identical recursive calls that are executed
more than once. For this reason, this technique resorts to evaluating the
recurrence in a bottom-up manner, saving intermediate results that are
used later on to compute the desired solution. This technique applies to
many combinatorial optimization problems to derive eﬃcient algorithms.
It is also used to improve the time complexity of the brute-force methods
to solve some of the NP-hard problems (see Chapter 10). For example, the
traveling salesman problem can be solved in time O(n
2
2
n
) using dynamic
programming, which is superior to the Θ(n!) bound of the obvious algo-
rithm that enumerates all possible tours. The two simple examples that
follow illustrate the essence of this design technique.
Example 7.1 One of the most popular examples used to introduce recursion
and induction is the problem of computing the Fibonacci sequence:
f
1
= 1, f
2
= 1, f
3
= 2, f
4
= 3, f
5
= 5, f
6
= 8, f
7
= 13, . . . .
Each number in the sequence 2, 3, 5, 8, 13, . . . is the the sum of the two preceding
203
204 Dynamic Programming
numbers. Consider the inductive deﬁnition of this sequence:
f(n) =

1 if n = 1 or n = 2
f(n −1) +f(n −2) if n ≥ 3.
This deﬁnition suggests a recursive procedure that looks like the following (as-
suming that the input is always positive).
1. procedure f (n)
2. if (n = 1) or (n = 2) then return 1
3. else return f(n −1) + f(n −2)
This recursive version has the advantages of being concise, easy to write and
debug and, most of all, its abstraction. It turns out that there is a rich class of
recursive algorithms and, in many instances, a complex algorithm can be written
succinctly using recursion. We have already encountered in the previous chapters
a number of eﬃcient algorithms that possess the merits of recursion. It should
not be thought, however, that the recursive procedure given above for computing
the Fibonacci sequence is an eﬃcient one. On the contrary, it is far from being
eﬃcient, as there are many duplicate recursive calls to the procedure. To see this,
just expand the recurrence a few times:
f(n) = f(n −1) +f(n −2)
= 2f(n −2) +f(n −3)
= 3f(n −3) + 2f(n −4)
= 5f(n −4) + 3f(n −5)
This leads to a huge number of identical calls. If we assume that computing
f(1) or f(2) requires a unit amount of time, then the time complexity of this
procedure can be stated as
T(n) =

1 if n = 1 or n = 2
T(n −1) +T(n −2) if n ≥ 3.
Clearly, the solution to this recurrence is T(n) = f(n), i.e., the time required to
compute f(n) is f(n) itself. It is well-known that f(n) ≈ Θ(φ
n
) for large n, where
φ = (1+
√
5)/2 ≈ 1.61803 is the golden ratio (see Example 2.20). In other words,
the running time required to compute f(n) is exponential in the value of n. An
obvious approach that reduces the time complexity drastically is to enumerate
the sequence bottom-up starting from f
1
until f
n
is reached. This takes Θ(n) time
and Θ(1) space; a substantial improvement.
The Longest Common Subsequence Problem 205
Example 7.2 As a similar example, consider computing the binomial coeﬃ-
cient

n
k

deﬁned recursively as

n
k

=

1 if k = 0 or k = n

n −1
k −1

+

n −1
k

if 0 < k < n.
Using the same argument as in Example 7.1, it can be shown that the time
complexity of computing

may proceed by constructing the Pascal
triangle row by row (see Fig. 2.1) and stopping as soon as the value of

n
k

has
been computed. The details will be left as an exercise (Exercise 7.3).
7.2 The Longest Common Subsequence Problem
A simple problem that illustrates the underlying principle of dynamic pro-
gramming is the following problem. Given two strings A and B of lengths
n and m, respectively, over an alphabet Σ, determine the length of the
longest subsequence that is common to both A and B. Here, a subsequence
of A = a
1
a
2
, . . . a
n
is a string of the form a
i
1
a
i
2
, . . . a
i
k
, where each i
j
is be-
tween 1 and n and 1 ≤ i
1
< i
2
< . . . < i
k
≤ n. For example, if Σ = ¦x, y, z¦,
A = zxyxyz and B = xyyzx, then xyy is a subsequence of length 3 of both
A and B. However, it is not the longest common subsequence of A and
B, since the string xyyz is also a common subsequence of length 4 of both
A and B. Since these two strings do not have a common subsequence of
length greater than 4, the length of the longest common subsequence of A
and B is 4.
206 Dynamic Programming
One way to solve this problem is to use the brute-force method: enu-
merate all the 2
n
subsequences of A, and for each subsequence determine
if it is also a subsequence of B in Θ(m) time. Clearly, the running time of
this algorithm is Θ(m2
n
), which is exponential.
In order to make use of the dynamic programming technique, we ﬁrst
ﬁnd a recursive formula for the length of the longest common subsequence.
Let A = a
1
a
2
, . . . a
n
and B = b
1
b
2
, . . . b
m
. Let L[i, j] denote the length of
a longest common subsequence of a
1
a
2
, . . . a
i
and b
1
b
2
, . . . b
j
. Note that i
or j may be zero, in which case one or both of a
1
a
2
, . . . a
i
and b
1
b
2
, . . . b
j
may be the empty string. Naturally, if i = 0 or j = 0, then L[i, j] = 0. The
following observation is easy to prove:
Observation 7.1 Suppose that both i and j are greater than 0. Then
• If a
i
= b
j
, L[i, j] = L[i −1, j −1] + 1.
• If a
i
= b
j
, L[i, j] = max¦L[i, j −1], L[i −1, j]¦.
The following recurrence for computing the length of the longest com-
mon subsequence of A and B follows immediately from Observation 7.1:
L[i, j] =

0 if i = 0 or j = 0
L[i −1, j −1] + 1 if i > 0, j > 0 and a
i
= b
j
max¦L[i, j −1], L[i −1, j]¦ if i > 0, j > 0 and a
i
= b
j
The algorithm
Using the technique of dynamic programming to solve the longest common
subsequence problem is now straightforward. We use an (n + 1) (m+ 1)
table to compute the values of L[i, j] for each pair of values of i and j,
0 ≤ i ≤ n and 0 ≤ j ≤ m. We only need to ﬁll the table L[0..n, 0..m]
row by row using the above formula. The method is formally described in
Algorithm lcs.
Algorithm lcs can easily be modiﬁed so that it outputs the longest
common subsequence. Clearly, the time complexity of the algorithm is
exactly the size of the table, Θ(nm), as ﬁlling each entry requires Θ(1) time.
The algorithm can easily be modiﬁed so that it requires only Θ(min¦m, n¦)
space. This implies the following theorem:
Theorem 7.1 An optimal solution to the longest common subsequence
problem can be found in Θ(nm) time and Θ(min¦m, n¦) space.
The Longest Common Subsequence Problem 207
Algorithm 7.1 lcs
Input: Two strings A and B of lengths n and m, respectively, over an alpha-
bet Σ.
Output: The length of the longest common subsequence of A and B.
1. for i ←0 to n
2. L[i, 0] ←0
3. end for
4. for j ←0 to m
5. L[0, j] ←0
6. end for
7. for i ←1 to n
8. for j ←1 to m
9. if a
i
= b
j
then L[i, j] ←L[i −1, j −1] + 1
10. else L[i, j] ← max¦L[i, j −1], L[i −1, j]¦
11. end if
12. end for
13. end for
14. return L[n, m]
Example 7.3 Figure 7.1 shows the result of applying Algorithm lcs on the
instance A =“xyxxzxyzxy” and B =“zxzyyzxxyxxz”.
0 1 2 3 4 5 6 7 8 9 10 11 12
0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 0 1 1 1 1 1 1 1 1 1 1 1
2 0 0 1 1 2 2 2 2 2 2 2 2 2
3 0 0 1 1 2 2 2 3 3 3 3 3 3
4 0 0 1 1 2 2 2 3 4 4 4 4 4
5 0 1 1 2 2 2 3 3 4 4 4 4 5
6 0 1 2 2 2 2 3 4 4 4 5 5 5
7 0 1 2 2 3 3 3 4 4 5 5 5 5
8 0 1 2 3 3 3 4 4 4 5 5 5 6
9 0 1 2 3 3 3 4 5 5 5 6 6 6
10 0 1 2 3 4 4 4 5 5 6 6 6 6
Fig. 7.1 An example of the longest common subsequence problem.
First, row 0 and column 0 are initialized to 0. Next, the entries are ﬁlled row
by row by executing Steps 5 and 6 exactly mn times. This generates the rest of
208 Dynamic Programming
the table. As shown in the table, the length of a longest common subsequence
is 6. One possible common subsequence is the string “xyxxxz” of length 6, which
can be constructed from the entries in the table in bold face.
7.3 Matrix Chain Multiplication
In this section, we study in detail another simple problem that reveals the
essence of dynamic programming. Suppose we want to compute the prod-
uct M
1
M
2
M
3
of three matrices M
1
, M
2
and M
3
of dimensions 210, 102
and 210, using the standard method of matrix multiplication. If we mul-
tiply M
1
and M
2
and then multiply the result by M
3
, the number of scalar
multiplications will be 2 10 2 +2 2 10 = 80. If, instead, we multiply
M
1
by the result of multiplying M
2
and M
3
, then the number of scalar mul-
tiplications becomes 10210+21010 = 400. Thus, carrying out the
multiplication M
1
(M
2
M
3
) costs ﬁve times the multiplication (M
1
M
2
)M
3
.
In general, the cost of multiplying a chain of n matrices M
1
M
2
. . . M
n
depends on the order in which the n−1 multiplications are carried out. That
order which minimizes the number of scalar multiplications can be found
in many ways. Consider for example the brute-force method that tries to
compute the number of scalar multiplications of every possible order. For
instance, if we have four matrices M
1
, M
2
, M
3
and M
4
, the algorithm will
try all the following ﬁve orderings:
(M
1
(M
2
(M
3
M
4
))),
(M
1
((M
2
M
3
)M
4
)),
((M
1
M
2
)(M
3
M
4
)),
((M
1
M
2
)M
3
)M
4
)),
((M
1
(M
2
M
3
))M
4
).
In general, the number of orderings is equal to the number of ways to
place parentheses to multiply the n matrices in every possible way. Let
f(n) be the number of ways to fully parenthesize a product of n matrices.
Suppose we want to perform the multiplication
(M
1
M
2
. . . M
k
) (M
k+1
M
k+2
. . . M
n
).
Then, there are f(k) ways to parenthesize the ﬁrst k matrices. For each
one of the f(k) ways, there are f(n−k) ways to parenthesize the remaining
Matrix Chain Multiplication 209
n − k matrices, for a total of f(k)f(n − k) ways. Since k can assume any
value between 1 and n −1, the overall number of ways to parenthesize the
n matrices is given by the summation
f(n) =
n−1
¸
k=1
f(k)f(n −k).
Observe that there is only one way to multiply two matrices and two ways
to multiply three matrices. That is, f(2) = 1 and f(3) = 2. In order for
the recurrence to make sense, we let f(1) = 1. It can be shown that
f(n) =
1
n

.
Since for each parenthesized expression, ﬁnding the number of scalar
multiplications costs Θ(n), it follows that the running time of the brute-
force method to ﬁnd the optimal way to multiply the n matrices is
Ω(4
n
/
√
n), which is impractical even for values of n of moderate size.
In the rest of this section, we derive a recurrence relation for the least
number of scalar multiplications, and then apply the dynamic programming
210 Dynamic Programming
technique to ﬁnd an eﬃcient algorithm for evaluating that recurrence. Ex-
tending the algorithm to ﬁnd the order of matrix multiplications is easy.
Since for each i, 1 ≤ i < n, the number of columns of matrix M
i
must
be equal to the number of rows of matrix M
i+1
, it suﬃces to specify the
number of rows of each matrix and the number of columns of the right-
most matrix M
n
. Thus, we will assume that we are given n+1 dimensions
r
1
, r
2
, . . . , r
n+1
, where r
i
and r
i+1
are, respectively, the number of rows
and columns in matrix M
i
, 1 ≤ i ≤ n. Henceforth, we will write M
i,j
to
denote the product of M
i
M
i+1
. . . M
j
. We will also assume that the cost of
multiplying the chain M
i,j
, denoted by C[i, j], is measured in terms of the
number of scalar multiplications. For a given pair of indices i and j with
1 ≤ i < j ≤ n, M
i,j
can be computed as follows. Let k be an index between
i + 1 and j. Compute the two matrices M
i,k−1
= M
i
M
i+1
. . . M
k−1
, and
M
k,j
= M
k
M
k+1
. . . M
j
. Then M
i,j
= M
i,k−1
M
k,j
. Clearly, the total cost
of computing M
i,j
in this way is the cost of computing M
i,k−1
plus the cost
of computing M
k,j
plus the cost of multiplying M
i,k−1
and M
k,j
, which is
r
i
r
k
r
j+1
. This leads to the following formula for ﬁnding that value of k
which minimizes the number of scalar multiplications required to perform
the matrix multiplication M
i
M
i+1
. . . M
j
.
C[i, j] = min
i<k≤j
¦C[i, k −1] +C[k, j] +r
i
r
k
r
j+1
¦. (7.1)
It follows that in order to ﬁnd the minimum number of scalar multi-
plications required to perform the matrix multiplication M
1
M
2
. . . M
n
, we
only need to solve the recurrence
C[1, n] = min
1<k≤n
¦C[1, k −1] +C[k, n] +r
1
r
k
r
n+1
¦.
However, as noted in Examples 7.1 and 7.2, this will lead to a huge number
of overlapping recursive calls, and hence solving the recurrence directly in
a top-down fashion will not result in an eﬃcient algorithm.
The dynamic programming algorithm
In what follows, we describe how the technique of dynamic programming
can be used to eﬃciently evaluate the above recurrence in time Θ(n
3
).
Consider Fig. 7.2, which illustrates the method on an instance consisting of
n = 6 matrices. In this ﬁgure, diagonal d is ﬁlled with the minimum costs
of multiplying various chains of d + 1 consecutive matrices. In particular,
diagonal 5 consists of exactly one entry which represents the minimum cost
Matrix Chain Multiplication 211
of multiplying the six matrices, which is the desired result. In diagonal 0,
each chain consists of one matrix only, and hence this diagonal is ﬁlled with
0’s. We ﬁll this triangular table with costs of multiplication diagonalwise,
starting at diagonal 0 and ending at diagonal 5. First, diagonal 0 is ﬁlled
with 0’s, as there are no scalar multiplications involved. Next, diagonal 1
is ﬁlled with the costs of multiplying two consecutive matrices. The rest
of the diagonals are ﬁlled using the formula stated above and the values
previously stored in the table. Speciﬁcally, to ﬁll diagonal d, we make use
of the values stored in diagonals 0, 1, 2, . . . , d −1.
1
2
3
4
5
6
d=1 d=2 d=3 d=4 d=5
C[1,1] C[1,2] C[1,3] C[1,4] C[1,5] C[1,6]
C[2,2] C[2,3] C[2,4] C[2,5] C[2,6]
C[3,3] C[3,4] C[3,6] C[3,5]
C[4,4] C[4,5] C[4,6]
C[5,5] C[5,6]
C[6,6]
j
d=0
Fig. 7.2 Illustration of matrix chain multiplication.
As an example, the computation of C[2, 5] is the minimum of the fol-
lowing three costs (see Fig. 7.2):
(1) The cost of computing M
2,2
(which is 0) plus the cost of computing
M
3,5
plus the cost of multiplying M
2,2
by M
3,5
.
(2) The cost of computing M
2,3
plus the cost of computing M
4,5
plus
the cost of multiplying M
2,3
by M
4,5
.
(3) The cost of computing M
2,4
plus the cost of computing M
5,5
(which
is 0) plus the cost of multiplying M
2,4
by M
5,5
.
To compute any other entry C[i, j] in the table other than the main
212 Dynamic Programming
diagonal, we do the following. First, we draw two directed vectors: one
from C[i, i] to C[i, j − 1] and one from C[i + 1, j] to C[j, j] (see Fig. 7.2).
Next, we compute the cost of multiplying each pair of matrices as we follow
the two arrows starting from the pair C[i, i] and C[i + 1, j] to the pair
C[i, j − 1] and C[j, j]. Finally, we select the minimum cost and store it in
C[i, j].
In general, multiplying a chain of n matrices gives rise to a triangu-
lar table of n rows and n columns similar to the one shown in Fig. 7.2.
The formal algorithm that produces such a table is given as Algorithm
matchain.
Algorithm 7.2 matchain
Input: An array r[1..n + 1] of positive integers corresponding to the
dimensions of a chain of n matrices, where r[1..n] are the number
of rows in the n matrices and r[n+1] is the number of columns in M
n
.
Output: The least number of scalar multiplications required to multiply
the n matrices
1. for i ←1 to n ¦Fill in diagonal d
0
¦
2. C[i, i] ←0
3. end for
4. for d←1 to n −1 ¦Fill in diagonals d
1
to d
n−1
¦
5. for i ←1 to n −d ¦Fill in entries in diagonal d
i
¦
6. j ←i +d
7. comment: The next three lines compute C[i, j]
8. C[i, j] ←∞
9. for k←i + 1 to j
10. C[i, j] ← min¦C[i, j], C[i, k−1] +C[k, j] +r[i]r[k]r[j +1]¦
11. end for
12. end for
13. end for
14. return C[1, n]
Step 1 ﬁlls diagonal 0 with 0’s. The execution of each iteration of the
for loop in Step 2 advances to the next diagonal. Each iteration of the
for loop in Step 3 advances to a new entry in that diagonal (each diagonal
contains n − d entries). Steps 6 to 8 compute entry C[i, j] using Eq. 7.1.
First, it is initialized to a very large value. Next, its value is chosen as the
minimum of d quantities corresponding to d multiplications of subchains as
explained for the case of the instance C[2, 5] described above and shown in
Fig. 7.2.
Matrix Chain Multiplication 213
Example 7.4 Figure 7.3 shows the result of applying Algorithm matchain
to ﬁnd the minimum number of scalar multiplication required to compute the
product of the following ﬁve matrices:
M
1
: 5 10, M
2
: 10 4, M
3
: 4 6, M
4
: 6 10, M
5
: 10 2.
Each entry C[i, j] of the upper triangular table is labeled with the minimum
number of scalar multiplications required to multiply the matrices M
i
M
i+1

. . . M
j
for 1 ≤ i ≤ j ≤ 5. The ﬁnal solution is C[1, 5] = 348.
C[1, 1] = 0 C[1, 2] = 200 C[1, 3] = 320 C[1, 4] = 620 C[1, 5] = 348
C[2, 2] = 0 C[2, 3] = 240 C[2, 4] = 640 C[2, 5] = 248
C[3, 3] = 0 C[3, 4] = 240 C[3, 5] = 168
C[4, 4] = 0 C[4, 5] = 120
C[5, 5] = 0
Fig. 7.3 An example of the matrix chain multiplication algorithm.
Finding the time and space complexites of Algorithm matchain is
straightforward. For some constant c > 0, the running time of the al-
gorithm is proportional to
n−1
¸
d=1
n−d
¸
i=1
d
¸
k=1
c =
cn
3
−cn
6
.
Hence, the time complexity of the algorithm is Θ(n
3
). Clearly, the work
space needed by the algorithm is dominated by that needed for the trian-
gular array, i.e., Θ(n
2
).
So far we have demonstrated an algorithm that computes the minimum
cost of multiplying a chain of matrices. It is not hard to modify the algo-
rithm so that it will also compute the actual order of matrix multiplications.
The following theorem summarizes the main result.
Theorem 7.2 The minimum number of scalar multiplications required to
multiply a chain of n matrices can be found in Θ(n
3
) time and Θ(n
2
) space.
Finally, we close this section by noting that, surprisingly, this problem
can be solved in time O(nlog n) (see the bibliographic notes).
214 Dynamic Programming
7.4 The Dynamic Programming Paradigm
Examples 7.1 and 7.2 and Secs. 7.2 and 7.3 provide an overview of the
dynamic programming algorithm design technique and its underlying prin-
ciple. The idea of saving solutions to subproblems in order to avoid their
recomputation is the basis of this powerful method. This is usually the
case in many combinatorial optimization problems in which the solution
can be expressed in the form of a recurrence whose direct solution causes
subinstances to be computed more than once.
An important observation about the working of dynamic programming
is that the algorithm computes an optimal solution to every subinstance of
the original instance considered by the algorithm. In other words, all the
table entries generated by the algorithm represent optimal solutions to the
subinstances considered by the algorithm.
For example, in Fig. 7.1, each entry L[i, j] is the length of a longest com-
mon subsequence for the subinstance obtained by taking the ﬁrst i letters
from the ﬁrst string and the ﬁrst j letters from the second string. Also, in
Fig. 7.2, each entry C[i, j] is the minimum number of scalar multiplications
needed to perform the product M
i
M
i+1
. . . M
j
. Thus, for example, the al-
gorithm that generates Fig. 7.2 not only computes the minimum number of
scalar multiplication for obtaining the product of the n matrices, but also
computes the minimum number of scalar multiplication of the product of
any sequence of consecutive matrices in M
1
M
2
. . . M
n
.
The above argument illustrates an important principle in algorithm de-
sign called the principle of optimality: Given an optimal sequence of deci-
sions, each subsequence must be an optimal sequence of decisions by itself.
We have already seen that the problems of ﬁnding the length of a longest
common subsequence and the problem of matrix chain multiplication can
be formulated in such a way that the principle of optimality applies. As
another example, let G = (V, E) be a directed graph and let π be a shortest
path from vertex s to vertex t, where s and t are two vertices in V . Suppose
that another vertex, say x ∈ V , is on this path. Then, it follows that the
portion of π from s to x must be a path of shortest length, and so is the
portion of π from x to t. This can trivially be proven by contradiction.
On the other hand, let π

be a simple path of longest length from s to t.
If vertex y ∈ V is on π

, then this does not mean, for example, that the
portion of π

from s to y is a longest simple path from s to y. This suggests
that dynamic programming may be used to ﬁnd a shortest path, but it is
The All-Pairs Shortest Path Problem 215
not obvious if it can be used to ﬁnd a longest simple path. In the case
of directed acyclic graphs, dynamic programming may be used to ﬁnd the
longest path between two given vertices (Exercise 7.34). Note that in this
case, all paths are simple.
7.5 The All-Pairs Shortest Path Problem
Let G = (V, E) be a directed graph in which each edge (i, j) has a non-
negative length l[i, j]. If there is no edge from vertex i to vertex j, then
l[i, j] = ∞. The problem is to ﬁnd the distance from each vertex to all other
vertices, where the distance from vertex x to vertex y is the length of a short-
est path from x to y. For simplicity, we will assume that V = ¦1, 2, . . . , n¦.
Let i and j be two diﬀerent vertices in V . Deﬁne d
k
i,j
to be the length
of a shortest path from i to j that does not pass through any vertex in
¦k + 1, k + 2, . . . , n¦. Thus, for example, d
0
i,j
= l[i, j], d
1
i,j
is the length of
a shortest path from i to j that does not pass through any vertex except
possibly vertex 1, d
2
i,j
is the length of a shortest path from i to j that does
not pass through any vertex except possibly vertex 1 or vertex 2 or both,
and so on. Then, by deﬁnition, d
n
i,j
is the length of a shortest path from i
to j, i.e., the distance from i to j. Given this deﬁnition, we can compute
d
k
i,j
recursively as follows.
d
k
i,j
=

l[i, j] if k = 0
min¦d
k−1
i,j
, d
k−1
i,k
+d
k−1
k,j
¦ if 1 ≤ k ≤ n.
The algorithm
The following algorithm, which is due to Floyd, proceeds by solving
the above recurrence in a bottom-up fashion. It uses n + 1 matrices
D
0
, D
1
, . . . , D
n
of dimension n n to compute the lengths of the short-
est constrained paths.
Initially, we set D
0
[i, i] = 0, D
0
[i, j] = l[i, j] if i = j and (i, j) is an edge
in G; otherwise D
0
[i, j] = ∞. We then make n iterations such that after
the kth iteration, D
k
[i, j] contains the value of a shortest length path from
vertex i to vertex j that does not pass through any vertex numbered higher
216 Dynamic Programming
than k. Thus, in the kth iteration, we compute D
k
[i, j] using the formula
D
k
[i, j] = min¦D
k−1
[i, j], D
k−1
[i, k] +D
k−1
[k, j]¦.
Example 7.5 Consider the directed graph shown in Fig. 7.4.
6
9
2 1
8
3 2
1
Fig. 7.4 An instance of the all pairs shortest path problem.
The matrices D
0
, D
1
, D
2
and D
3
are
D
0
=
¸
0 2 9
8 0 6
1 ∞ 0
¸
D
1
=
¸
0 2 9
8 0 6
1 3 0
¸
D
2
=
¸
0 2 8
8 0 6
1 3 0
¸
D
3
=
¸
0 2 8
7 0 6
1 3 0
¸
.
The ﬁnal computed matrix D
3
holds the desired distances.
An important observation is that in the kth iteration, both the kth row
and kth column are not changed. Therefore, we can perform the compu-
tation with only one copy of the D matrix. An algorithm to perform this
computation using only one n n matrix is given as Algorithm floyd.
Algorithm 7.3 floyd
Input: An n n matrix l[1..n, 1..n] such that l[i, j] is the length of the edge
(i, j) in a directed graph G = (¦1, 2, . . . , n¦, E).
Output: A matrix D with D[i, j] = the distance from i to j.
1. D←l ¦copy the input matrix l into D¦
2. for k←1 to n
3. for i ←1 to n
4. for j ←1 to n
5. D[i, j] = min¦D[i, j], D[i, k] +D[k, j]¦
6. end for
7. end for
8. end for
The Knapsack Problem 217
Clearly, the running time of the algorithm is Θ(n
3
) and its space com-
plexity is Θ(n
2
).
7.6 The Knapsack Problem
The knapsack problem can be deﬁned as follows. Let U = ¦u
1
, u
2
, . . . , u
n
¦
be a set of n items to be packed in a knapsack of size C. For 1 ≤ j ≤ n,
let s
j
and v
j
be the size and value of the jth item, respectively, where C
and s
j
, v
j
, 1 ≤ j ≤ n, are all positive integers. The objective is to ﬁll the
knapsack with some items from U whose total size is at most C and such
that their total value is maximum. Assume without loss of generality that
the size of each item does not exceed C. More formally, given U of n items,
we want to ﬁnd a subset S ⊆ U such that
¸
u
i
∈S
v
i
is maximized subject to the constraint
¸
u
i
∈S
s
i
≤ C.
This version of the knapsack problem is sometimes referred to in the liter-
ature as the 0/1 knapsack problem. This is because the knapsack cannot
contain more than one item of the same type. Another version of the prob-
lem in which the knapsack may contain more than one item of the same
type is discussed in Exercise 7.27.
We derive a recursive formula for ﬁlling the knapsack as follows. Let
V [i, j] denote the value obtained by ﬁlling a knapsack of size j with items
taken from the ﬁrst i items ¦u
1
, u
2
, . . . , u
i
¦ in an optimal way. Here the
range of i is from 0 to n and the range of j is from 0 to C. Thus, what we
seek is the value V [n, C]. Obviously, V [0, j] is 0 for all values of j, as there
is nothing in the knapsack. On the other hand, V [i, 0] is 0 for all values of
i since nothing can be put in a knapsack of size 0. For the general case,
when both i and j are greater than 0, we have the following observation,
which is easy to prove:
Observation 7.2 V [i, j] is the maximum of the following two quantities:
218 Dynamic Programming
• V [i − 1, j]: The maximum value obtained by ﬁlling a knapsack of
size j with items taken from ¦u
1
, u
2
, . . . , u
i−1
¦ only in an optimal
way.
• V [i − 1, j − s
i
] + v
i
: The maximum value obtained by ﬁlling a
knapsack of size j −s
i
with items taken from ¦u
1
, u
2
, . . . , u
i−1
¦ in
an optimal way plus the value of item u
i
. This case applies only if
j ≥ s
i
and it amounts to adding item u
i
to the knapsack.
Observation 7.2 implies the following recurrence for ﬁnding the value in
an optimal packing:
V [i, j] =

0 if i = 0 or j = 0
V [i −1, j] if j < s
i
max¦V [i −1, j], V [i −1, j −s
i
] +v
i
¦ if i > 0 and j ≥ s
i
.
The algorithm
Using dynamic programming to solve this integer programming problem is
now straightforward. We use an (n + 1) (C + 1) table to evaluate the
values of V [i, j]. We only need to ﬁll the table V [0..n, 0..C] row by row
using the above formula. The method is formally described in Algorithm
knapsack.
Algorithm 7.4 knapsack
Input: A set of items U = ¦u
1
, u
2
, . . . , u
n
¦ with sizes s
1
, s
2
, . . . , s
n
and
values v
1
, v
2
, . . . , v
n
and a knapsack capacity C.
Output: The maximum value of the function
¸
u
i
∈S
v
i
subject to
¸
u
i
∈S
s
i
≤ C for some subset of items S ⊆ U.
1. for i ←0 to n
2. V [i, 0] ←0
3. end for
4. for j ←0 to C
5. V [0, j] ←0
6. end for
7. for i ←1 to n
8. for j ←1 to C
9. V [i, j] ←V [i −1, j]
10. if s
i
≤ j then V [i, j] ← max¦V [i, j], V [i −1, j −s
i
] +v
i
¦
11. end for
12. end for
13. return V [n, C]
The Knapsack Problem 219
Clearly, the time complexity of the algorithm is exactly the size of the
table, Θ(nC), as ﬁlling each entry requires Θ(1) time. Algorithm knap-
sack can easily be modiﬁed so that it outputs the items packed in the
knapsack as well. It can also be easily modiﬁed so that it requires only
Θ(C), as only the last computed row is needed for ﬁlling the current row.
This implies the following theorem:
Theorem 7.3 An optimal solution to the Knapsack problem can be
found in Θ(nC) time and Θ(C) space.
Note that the time bound, as stated in the above theorem, is not poly-
nomial in the input size. Therefore, the algorithm is considered to be
exponential in the input size. For this reason, it is referred to as a pseu-
dopolynomial time algorithm, as the running time is polynomial in the input
value.
Example 7.6 Suppose that we have a knapsack of capacity 9, which we want
to pack with items of four diﬀerent sizes 2, 3, 4 and 5 and values 3, 4, 5 and 7,
respectively. Our goal is to pack the knapsack with as many items as possible in a
way that maximizes the total value without exceeding the knapsack capacity. We
proceed to solve this problem as follows. First, we prepare an empty rectangular
table with ﬁve rows numbered 0 to 4 and 10 columns labeled 0 through 9. Next,
we initialize the entries in column 0 and row 0 with the value 0. Filling row 1 is
straightforward: V [1, j] = 3, the value of the ﬁrst item, if and only if j ≥ 2, the
size of the ﬁrst item. Each entry V [2, j] in the second column has two possibilities.
The ﬁrst possibility is to set V [2, j] = V [1, j], which amounts to putting the ﬁrst
item in the knapsack. The second possibility is to set V [2, j] = V [1, j − 3] + 4,
which amounts to adding the second item so that it either contains the second
item only or both the ﬁrst and second items. Of course, adding the second item
is possible only if j ≥ 3. Continuing this way, rows 3 and 4 are ﬁlled to obtain
the table shown in Fig. 7.5.
0 1 2 3 4 5 6 7 8 9
0 0 0 0 0 0 0 0 0 0 0
1 0 0 3 3 3 3 3 3 3 3
2 0 0 3 4 4 7 7 7 7 7
3 0 0 3 4 4 7 8 9 9 12
4 0 0 3 4 5 7 8 10 11 12
Fig. 7.5 An example of the algorithm for the knapsack problem.
220 Dynamic Programming
The ith entry of column 9, that is, V [i, 9] contains the maximum value we
can get by ﬁlling the knapsack using the ﬁrst i items. Thus, an optimal packing
is found in the last entry of the last column and is achieved by packing items 3
and 4. There is also another optimal solution, which is packing items 1, 2 and
3. This packing corresponds to entry V [3, 9] in the table, which is the optimal
packing before the fourth item was considered.
7.7 Exercises
7.1. We have deﬁned the dynamic programming paradigm in such a way
that it encompasses all algorithms that solve a problem by breaking it
down into smaller subproblems, saving the solution to each subproblem
and using these solutions to compute an optimal solution to the main
problem. Which of the following algorithms can be classiﬁed as dynamic
programming algorithms?
(a) Algorithm linearsearch.
(b) Algorithm insertionsort.
(c) Algorithm bottomupsort.
(d) Algorithm mergesort.
7.2. Give an eﬃcient algorithm to compute f(n), the nth number in the
Fibonacci sequence (see Example 7.1). What is the time complexity of
your algorithm? Is it an exponential algorithm? Explain.
7.3. Give an eﬃcient algorithm to compute the binomial coeﬃcient

n
k

(see
Example 7.2). What is the time complexity of your algorithm? Is it an
exponential algorithm? Explain.
7.4. Prove Observation 7.1.
7.5. Use Algorithm lcs to ﬁnd the length of a longest common subsequence
of the two strings A = “xzyzzyx” and B = “zxyyzxz”. Give one longest
common subsequence.
7.6. Show how to modify Algorithm lcs so that it outputs a longest common
subsequence as well.
7.7. Show how to modify Algorithm lcs so that it requires only Θ(min¦m, n¦)
space.
7.8. In Sec. 7.3, it was shown that the number of ways to fully parenthesize
Exercises 221
n matrices is given by the summation
f(n) =
n−1
¸
k=1
f(k)f(n −k).
Show that the solution to this recurrence is
f(n) =
1
n

be a longest simple path from s to t that passes by another
vertex y. Show that the portion of the path from s to y is not
necessarily a longest path from s to y.
7.15. Run the all-pairs shortest path algorithm on the weighted directed graph
shown in Fig. 7.7.
6 9
2
1
4
1
4 3
2
7
4
1
Fig. 7.7 An instance of the all-pairs shortest path problem.
7.16. Use the all-pairs shortest path algorithm to compute the distance matrix
for the directed graph with the lengths of the edges between all pairs of
vertices are as given by the matrix
(a)

0 1 ∞ 2
2 0 ∞ 2
∞ 9 0 4
8 2 3 0
¸
¸
¸
(b)

0 2 4 6
2 0 1 2
5 9 0 1
9 ∞ 2 0
¸
¸
¸
.
7.17. Give an example of a directed graph that contains some edges with neg-
ative costs and yet the all-pairs shortest path algorithm gives the correct
distances.
Exercises 223
7.18. Give an example of a directed graph that contains some edges with neg-
ative costs such that the all-pairs shortest path algorithm fails to give
the correct distances.
7.19. Show how to modify the all-pairs shortest path algorithm so that it
detects negative-weight cycles (A negative-weight cycle is a cycle whose
total length is negative).
7.20. Prove Observation 7.2.
7.21. Solve the following instance of the knapsack problem. There are four
items of sizes 2, 3, 5, and 6 and values 3, 4, 5, and 7, and the knapsack
capacity is 11.
7.22. Solve the following instance of the knapsack problem. There are ﬁve
items of sizes 3, 5, 7, 8 and 9 and values 4, 6, 7, 9 and 10, and the
knapsack capacity is 22.
7.23. Explain what would happen when running the knapsack algorithm on
an input in which one item has negative size.
7.24. Show how to modify Algorithm knapsack so that it requires only Θ(C)
space, where C is the knapsack capacity.
7.25. Show how to modify Algorithm knapsack so that it outputs the items
packed in the knapsack as well.
7.26. In order to lower the prohibitive running time of the knapsack problem,
which is Θ(nC), we may divide C and all the s
i
’s by a large number K
and take the ﬂoor. That is, we may transform the given instance into
a new instance with capacity ]C/K and item sizes ]s
i
/K, 1 ≤ i ≤ n.
Now, we apply the algorithm for the knapsack discussed in Sec. 7.6. This
technique is called scaling and rounding (see Sec. 15.6). What will be the
running time of the algorithm when applied to the new instance? Give
a counterexample to show that scaling and rounding does not always
result in an optimal solution to the original instance.
7.27. Another version of the knapsack problem is to let the set U contain a
set of types of items, and the objective is to ﬁll the knapsack with any
number of items of each type in order to maximize the total value without
exceeding the knapsack capacity. Assume that there is an unlimited
number of items of each type. More formally, let T = ¦t
1
, t
2
, . . . , t
n
¦ be
a set of n types of items, and C the knapsack capacity. For 1 ≤ j ≤ n,
let s
j
and v
j
be, respectively, the size and value of the items of type j.
Find a set of nonnegative integers x
1
, x
2
, . . . , x
n
such that
n
¸
i=1
x
i
v
i
224 Dynamic Programming
is maximized subject to the constraint
n
¸
i=1
x
i
s
i
≤ C.
x
1
, x
2
, . . . , x
n
are nonnegative integers.
Note that x
j
= 0 means that no item of the jth type is packed in the
knapsack. Rewrite the dynamic programming algorithm for this version
of the knapsack problem.
7.28. Solve the following instance of the version of the knapsack problem de-
scribed in Exercise 7.27. There are ﬁve types of items with sizes 2, 3, 5
and 6 and values 4, 7, 9 and 11, and the knapsack capacity is 8.
7.29. Show how to modify the knapsack algorithm discussed in Exercise 7.27
so that it computes the number of items packed from each type.
7.30. Consider the money change problem. We have a currency system that
has n coins with values v
1
, v
2
, . . . , v
n
, where v
1
= 1, and we want to
pay change of value y in such a way that the total number of coins is
minimized. More formally, we want to minimize the quantity
n
¸
i=1
x
i
subject to the constraint
n
¸
i=1
x
i
v
i
= y.
Here, x
1
, x
2
, . . . , x
n
are nonnegative integers (so x
i
may be zero).
(a) Give a dynamic programming algorithm to solve this problem.
(b) What are the time and space complexities of your algorithm?
(c) Can you see the resemblance of this problem to the version of the
knapsack problem discussed in Exercise 7.27? Explain.
7.31. Apply the algorithm in Exercise 7.30 to the instance v
1
= 1, v
2
= 5, v
3
=
7, v
4
= 11 and y = 20.
7.32. Let G = (V, E) be a directed graph with n vertices. G induces a relation
R on the set of vertices V deﬁned by: u R v if and only if there is a
directed edge from u to v, i.e., if and only if (u, v) ∈ E. Let M
R
be the
adjacency matrix of G, i.e., M
R
is an nn matrix satisfying M
R
[u, v] = 1
if (u, v) ∈ E and 0 otherwise. The reﬂexive and transitive closure of M
R
,
denoted by M
∗
R
, is deﬁned as follows. For u, v ∈ V , if u = v or there
is a path in G from u to v, then M
∗
R
[u, v] = 1 and 0 otherwise. Give
a dynamic programming algorithm to compute M
∗
R
for a given directed
Exercises 225
graph. (Hint: You only need a slight modiﬁcation of Floyd’s algorithm
for the all-pairs shortest path problem).
7.33. Let G = (V, E) be a directed graph with n vertices. Deﬁne the n n
distance matrix D as follows. For u, v ∈ V , D[u, v] = d if and only if
the length of the shortest path from u to v measured in the number of
edges is exactly d. For example, for any v ∈ V , D[v, v] = 0 and for
any u, v ∈ V D[u, v] = 1 if and only if (u, v) ∈ E. Give a dynamic
programming algorithm to compute the distance matrix D for a given
directed graph. (Hint: Again, you only need a slight modiﬁcation of
Floyd’s algorithm for the all-pairs shortest path problem).
7.34. Let G = (V, E) be a directed acyclic graph (dag) with n vertices. Let
s and t be two vertices in V such that the indegree of s is 0 and the
outdegree of t is 0. Give a dynamic programming algorithm to compute
a longest path in G from s to t. What is the time complexity of your
algorithm?
7.35. Give a dynamic programming algorithm for the traveling salesman prob-
lem: Given a set of n cities with their intercity distances, ﬁnd a tour of
minimum length. Here, a tour is a cycle that visits each city exactly
once. What is the time complexity of your algorithm? This problem can
be solved using dynamic programming in time O(n
2
2
n
) (see the biblio-
graphic notes).
7.36. Let P be a convex polygon with n vertices (see Sec. 18.2). A chord in
P is a line segment that connects two nonadjacent vertices in P. The
problem of triangulating a convex polygon is to partition the polygon
into n −2 triangles by drawing n −3 chords inside P. Figure 7.8 shows
two possible triangulations of the same convex polygon.
Fig. 7.8 Two triangulations of the same convex polygon.
(a) Show that the number of ways to triangulate a convex polygon
with n vertices is the same as the number of ways to multiply n−1
matrices.
(b) A minimum weight triangulation is a triangulation in which the
sum of the lengths of the n−3 chords is minimum. Give a dynamic
programming algorithm for ﬁnding a minimum weight triangulation
of a convex polygon with n vertices. (Hint: This problem is very
similar to the matrix chain multiplication covered in Sec. 7.3).
226 Dynamic Programming
7.8 Bibliographic notes
Dynamic programming was ﬁrst popularized in the book by Bellman (1957).
Other books in this area include Bellman and Dreyfus (1962), Dreyfus
(1977) and Nemhauser (1966). Two general survey papers by Brown (1979)
and Held and Karp (1967) are highly recommended. The all-pairs short-
est paths algorithm is due to Floyd (1962). Matrix chain multiplication is
described in Godbole (1973). An O(nlog n) algorithm to solve this prob-
lem can be found in Hu and Shing (1980, 1982, 1984). The one and two
dimensional knapsack problems have been studied extensively; see for ex-
ample Gilmore (1977), Gilmore and Gomory (1966) and Hu (1969). Held
and Karp (1962) gave an O(n
2
2
n
) dynamic programming algorithm for the
traveling salesman problem. This algorithm also appears in Horowitz and
Sahni (1978).
PART 3
First-Cut Techniques
227
228
229
When a solution to a problem is sought, perhaps the ﬁrst strategy that
comes to one’s mind is the greedy method. If the problem involves graphs,
then one might consider traversing the graph, visiting its vertices and per-
forming some actions depending on a decision made at that point. The
technique used to solve that problem is usually speciﬁc to the problem itself.
A common characteristic of both greedy algorithms and graph traversal is
that they are fast, as they involve making local decisions.
A graph traversal algorithm might be viewed as a greedy algorithm and
vice-versa. In graph traversal techniques, the choice of the next vertex to
be examined is restricted to the set of neighbors of the current node. This
is in contrast to examining a bigger neighborhood, clearly a simple greedy
strategy. On the other hand, a greedy algorithm can also be viewed as a
graph traversal of a particular graph. For any greedy algorithm, there is
an implicit directed acyclic graph (dag) each of whose nodes stands for a
state in that greedy computation. An intermediate state represents some
decisions that were already taken in a greedy fashion, while others remain
to be determined. In that dag, an edge from vertex u to vertex v exists only
if in the greedy method the algorithm’s state represented by v is arrived at
from that represented by vertex u as a consequence of one decision by the
greedy algorithm.
Although these techniques tend to be applied as initial solutions, they
rarely remain as the providers of optimal solutions. Their contribution
consequently is one of providing an initial solution that sets the stage for
careful examination of the speciﬁc properties of the problem.
In Chapter 8, we study in detail some algorithms that give optimal solu-
tions to well-known problems in computer science and engineering. The two
famous problems of the single-source shortest path, and ﬁnding a minimum
cost spanning tree in an undirected graph are representative of those prob-
lems for which the greedy strategy results in an optimal solution. Other
problems, like Huﬀman code, will also be covered in this chapter.
Chapter 9 is devoted to graph traversals (depth-ﬁrst search and breadth-
ﬁrst search) that are useful in solving many problems, especially graph and
geometric problems.
230
Chapter 8
The Greedy Approach
8.1 Introduction
As in the case of dynamic programming algorithms, greedy algorithms are
usually designed to solve optimization problems in which a quantity is to
be minimized or maximized. However, unlike dynamic programming algo-
rithms, greedy algorithms typically consist of an iterative procedure that
tries to ﬁnd a local optimal solution. In some instances, these local op-
timal solutions translate to global optimal solutions. In others, they fail
to give optimal solutions. A greedy algorithm makes a correct guess on
the basis of little calculation without worrying about the future. Thus, it
builds a solution step by step. Each step increases the size of the partial
solution and is based on local optimization. The choice made is that which
produces the largest immediate gain while maintaining feasibility. Since
each step consists of little work based on a small amount of information,
the resulting algorithms are typically eﬃcient. The hard part in the design
of a greedy algorithm is proving that the algorithm does indeed solve the
problem it is designed for. This is to be contrasted with recursive algo-
rithms that usually have very simple inductive proofs. In this chapter, we
will study some of the most prominent problems for which the greedy strat-
egy works, i.e., gives an optimal solution: the single-source shortest path
problem, minimum cost spanning trees (Prim’s and Kruskal’s algorithms)
and Huﬀman codes. We will postpone those greedy algorithms that give
suboptimal solutions to Chapter 15. The exercises contain some problems
for which the greedy strategy works (e.g. Exercises 8.3, 8.10 and 8.34) and
others for which the greedy method fails to give the optimal solution on
231
232 The Greedy Approach
some instances (e.g. Exercises 8.7, 8.8, 8.9 and 8.12). The following is a
simple example of a problem for which the greedy strategy works.
Example 8.1 Consider the fractional knapsack problem deﬁned as follows.
Given n items of sizes s
1
, s
2
, . . . , s
n
, and values v
1
, v
2
, . . . , v
n
and size C, the
knapsack capacity, the objective is to ﬁnd nonnegative real numbers x
1
, x
2
, . . . , x
n
that maximize the sum
n
¸
i=1
x
i
v
i
subject to the constraint
n
¸
i=1
x
i
s
i
≤ C.
This problem can easily be solved using the following greedy strategy. For
each item compute y
i
= v
i
/s
i
, the ratio of its value to its size. Sort the items by
decreasing ratio, and ﬁll the knapsack with as much as possible from the ﬁrst item,
then the second, and so forth. This problem reveals many of the characteristics of
a greedy algorithm discussed above: The algorithm consists of a simple iterative
procedure that selects that item which produces the largest immediate gain while
maintaining feasibility.
8.2 The Shortest Path Problem
Let G = (V, E) be a directed graph in which each edge has a nonnegative
length, and a distinguished vertex s called the source. The single-source
shortest path problem, or simply the shortest path problem, is to determine
the distance from s to every other vertex in V , where the distance from
vertex s to vertex x is deﬁned as the length of a shortest path from s
to x. For simplicity, we will assume that V = ¦1, 2, . . . , n¦ and s = 1.
This problem can be solved using a greedy technique known as Dijkstra’s
algorithm. Initially, the set of vertices is partitioned into two sets X = ¦1¦
and Y = ¦2, 3, . . . , n¦. The intention is that X contains the set of vertices
whose distance from the source has already been determined. At each step,
we select a vertex y ∈ Y whose distance from the source vertex has already
been found and move it to X. Associated with each vertex y in Y is a
label λ[y], which is the length of a shortest path that passes only through
vertices in X. Once a vertex y ∈ Y is moved to X, the label of each vertex
w ∈ Y that is adjacent to y is updated indicating that a shorter path to w
The Shortest Path Problem 233
via y has been discovered. Throughout this section, for any vertex v ∈ V ,
δ[v] will denote the distance from the source vertex to v. As will be shown
later, at the end of the algorithm, δ[v] = λ[v] for each vertex v ∈ V . A
sketch of the algorithm is given below.
1. X ←¦1¦; Y ←V −¦1¦
2. For each vertex v ∈ Y if there is an edge from 1 to v then let λ[v]
(the label of v) be the length of that edge; otherwise let λ[v] = ∞.
Let λ[1] = 0.
3. while Y ,= ¦¦
4. Let y ∈ Y be such that λ[y] is minimum.
5. move y from Y to X.
6. update the labels of those vertices in Y that are adjacent to y.
7. end while
1
2
3
6
4
5
1
12
15
4
9 13
4
3
5
(e)
1
0
8
4
17
13
1
2
3
6
4
5
1
12
15
4
9 13
4
3
5
(f)
1
0
8
4
17
13
1
2
3
6
4
5
1
12
15
4
9 13
4
3
5
(c)
1
0
8
4
19
17
1
2
3
6
4
5
1
12
15
4
9 13
4
3
5
(d)
1
0
8
4
19
13
1
2
3
6
4
5
1
12
15
4
9 13
4
3
5
(b)
1
0
10
4
1
2
3
6
4
5
1
12
15
4
9 13
4
3
5
(a)
1
0
12
Fig. 8.1 An example of Dijkstra’s algorithm.
234 The Greedy Approach
Example 8.2 To see how the algorithm works, consider the directed graph
shown in Fig. 8.1(a). The ﬁrst step is to label each vertex v with λ[v] =
length[1, v]. As shown in the ﬁgure, vertex 1 is labeled with 0, and vertices 2
and 3 are labeled with 1 and 12 since length[1, 2] = 1 and length[1, 3] = 12. All
other vertices are labeled with ∞ since there are no edges from the source vertex
to these vertices. Initially X = ¦1¦ and Y = ¦2, 3, 4, 5, 6¦. In the ﬁgure, those
vertices to the left of the dashed line belong to X, and the others belong to Y . In
Fig. 8.1(a), we note that λ[2] is the smallest among all vertices’ labels in Y , and
hence it is moved to X indicating that the distance to vertex 2 has been found.
To ﬁnish processing vertex 2, the labels of its neighbors 3 and 4 are inspected to
see if there are paths that pass through 2 and are shorter than their old paths. In
this case, we say that we update the labels of the vertices adjacent to 2. As shown
in the ﬁgure, the path from 1 to 2 to 3 is shorter than the path from 1 to 3, and
thus λ[3] is changed to 10, which is the length of the path that passes through 2.
Similarly, λ[4] is changed to 4 since now there is a ﬁnite path of length 4 from 1
to 4 that passes through vertex 2. These updates are shown in Fig. 8.1(b). The
next step is to move that vertex with minimum label, namely 4, to X and update
the labels of its neighbors in Y as shown in Fig. 8.1(c). In this ﬁgure we notice
that the labels of vertices 5 and 6 became ﬁnite and λ[3] is lowered to 8. Now,
vertex 3 has a minimum label, so it is moved to X and λ[5] is updated accord-
ingly as shown in Fig. 8.1(d). Continuing in this way, the distance to vertex 5 is
found and thus it is moved to X as shown in Fig. 8.1(e). As shown in Fig. 8.1(f),
vertex 6 is the only vertex remaining in Y and hence its label coincides with the
length of its distance from 1. In Fig. 8.1(f), the label of each vertex represents
its distance from the source vertex.
Implementation of the shortest path algorithm
A more detailed description of the algorithm is given in Algorithm dijk-
stra.
We will assume that the input graph is represented by adjacency lists,
and the length of edge (x, y) is stored in the vertex for y in the adjacency
list for x. For instance, the directed graph shown in Fig. 8.1 is represented
as shown in Fig. 8.2. We will also assume that the length of each edge in
E is nonnegative. The two sets X and Y will be implemented as boolean
vectors X[1..n] and Y [1..n]. Initially, X[1] = 1 and Y [1] = 0, and for all
i, 2 ≤ i ≤ n, X[i] = 0 and Y [i] = 1. Thus, the operation X←X ∪ ¦y¦
is implemented by setting X[y] to 1, and the operation Y ←Y − ¦y¦ is
implemented by setting Y [y] to 0.
The Shortest Path Problem 235
Algorithm 8.1 dijkstra
Input: A weighted directed graph G = (V, E), where V = ¦1, 2, . . . , n¦.
Output: The distance from vertex 1 to every other vertex in G.
1. X = ¦1¦; Y ←V −¦1¦; λ[1] ←0
2. for y←2 to n
3. if y is adjacent to 1 then λ[y] ←length[1, y]
4. else λ[y] ←∞
5. end if
6. end for
7. for j ←1 to n −1
8. Let y ∈ Y be such that λ[y] is minimum
9. X←X ∪ ¦y¦ ¦add vertex y to X¦
10. Y ←Y −¦y¦ ¦delete vertex y from Y ¦
11. for each edge (y, w)
12. if w ∈ Y and λ[y] + length[y, w] < λ[w] then
13. λ[w] ←λ[y] + length[y, w]
14. end for
15. end for
1
2
3
4
5
3
5 3
6
2
3 4
5
6
6
1 12
3 9
5
4 13 15
4
Fig. 8.2 Directed graph representation for the shortest path algorithm.
Correctness
Lemma 8.1 In Algorithm dijkstra, when a vertex y is chosen in Step 5,
if its label λ[y] is ﬁnite then λ[y] = δ[y].
Proof. By induction on the order in which vertices leave the set Y and
enter X. The ﬁrst vertex to leave is 1 and we have λ[1] = δ[1] = 0. Assume
that the statement is true for all vertices which left Y before y. Since λ[y]
is ﬁnite, there must exists a path from 1 to y whose length is λ[y]. Now, we
show that λ[y] ≤ δ[y]. Let π = 1, . . . , x, w, . . . , y be a shortest path from 1
236 The Greedy Approach
to y, where x is the rightmost vertex to leave Y before y (see Fig. 8.3). We
have
λ[y] ≤ λ[w] since y left Y before w
≤ λ[x] +length(x, w) by the algorithm
= δ[x] +length(x, w) by induction
= δ[w] since π is a path of shortest length
≤ δ[y] since π is a path of shortest length.
⁄
It will be left as an exercise to show that the above proof is based on
the assumption that all edge lengths are nonnegative (Exercise 8.20).
a
n
j
A
1
Fig. 8.3 Proof of correctness of Algorithm dijkstra.
Time complexity
The time complexity of the algorithm is computed as follows. Step 1 costs
Θ(n) time. Steps 2 and 3 cost Θ(n) and O(n), respectively. The time
taken by Step 5 to search for the vertex with minimum label is Θ(n). This
is because the algorithm has to inspect each entry in the vector representing
the set Y . Since it is executed n − 1 times, the overall time required by
Step 5 is Θ(n
2
). Steps 6 and 7 cost Θ(1) time per iteration for a total of
Θ(n) time. The for loop in Step 8 is executed m times throughout the
algorithm, where m = [E[. This is because each edge (y, w) is inspected
exactly once by the algorithm. Hence, the overall time required by steps 8
and 9 is Θ(m). It follows that the time complexity of the algorithm is
Θ(m+n
2
) = Θ(n
2
).
The Shortest Path Problem 237
Theorem 8.1 Given a directed graph G with nonnegative weights on its
edges and a source vertex s, Algorithm dijkstra ﬁnds the length of the
distance from s to every other vertex in Θ(n
2
) time.
Proof. Lemma 8.1 establishes the correctness of the algorithm and the
time complexity follows from the above discussion. ⁄
8.2.1 A linear time algorithm for dense graphs
Now we are ready to make a major improvement to Algorithm dijkstra in
order to lower its Θ(n
2
) time complexity to O(mlog n) for graphs in which
m = o(n
2
). We will also improve it further, so that in the case of dense
graphs it runs in time linear in the number of edges.
The basic idea is to use the min-heap data structure (see Sec. 4.2) to
maintain the vertices in the set Y so that the vertex y in Y closest to a
vertex in V −Y can be extracted in O(log n) time. The key associated with
each vertex v is its label λ[v]. The ﬁnal algorithm is shown as Algorithm
shortestpath.
It should be emphasized that the input to the algorithm is the adjacency
lists; otherwise we will end up with a Θ(n
2
) time algorithm if the adjacency
matrix is used. Each vertex y ∈ Y is assigned a key which is the cost of
the edge connecting 1 to y if it exists; otherwise that key is set to ∞. The
heap H initially contains all vertices adjacent to vertex 1. Each iteration
of the for loop in Step 6 starts by extracting that vertex y with minimum
key. The key of each vertex w in Y adjacent to y is then updated. Next,
if w is not in the heap, then it is inserted; otherwise it is sifted up, if
necessary. The function H
−1
(w) returns the position of w in H. This can
be implemented by simply having an array that has for its jth entry the
position of vertex j in the heap (recall that the heap is implemented as
an array H[1..n]). The running time is dominated by the heap operations.
There are n − 1 deletemin operations, n − 1 insert operations and at
most m − n + 1 siftup operations. Each heap operation takes O(log n)
time, which results in O(mlog n) time in total.
Recall that a d-heap is essentially a generalization of a binary heap
in which each internal node in the tree has at most d children in-
stead of 2, where d is a number that can be arbitrarily large (see Ex-
ercise 4.21). If we use a d-heap, the running time is improved as fol-
lows. Each deletemin operation takes O(d log
d
n) time, and each in-
238 The Greedy Approach
Algorithm 8.2 shortestpath
Input: A weighted directed graph G = (V, E), where V = ¦1, 2, . . . , n¦.
Output: The distance from vertex 1 to every other vertex in G.
Assume that we have an empty heap H at the beginning.
1. Y ←V −¦1¦; λ[1] ←0; key(1)←λ[1]
2. for y←2 to n
3. if y is adjacent to 1 then
4. λ[y] ←length[1, y]
5. key(y)←λ[y]
6. insert(H, y)
7. else
8. λ[y] ←∞
9. key(y)←λ[y]
10. end if
11. end for
12. for j ←1 to n −1
13. y←deletemin(H)
14. Y ←Y −¦y¦ ¦delete vertex y from Y ¦
15. for each vertex w ∈ Y that is adjacent to y
16. if λ[y] + length[y, w] < λ[w] then
17. λ[w] ←λ[y] + length[y, w]
18. key(w)←λ[w]
19. end if
20. if w / ∈ H then insert(H, w)
21. else siftup(H, H
−1
(w))
22. end if
23. end for
24. end for
sert or siftup operation requires O(log
d
n) time. Thus, the total running
time is O(nd log
d
n + mlog
d
n). If we choose d = 2 +m/n|, the time
bound is O(mlog
2+m/n
n). If m ≥ n
1+
for some > 0 that is not too
small, i.e., the graph is dense, then the running time is
O(mlog
2+m/n
n) = O(mlog
2+n

n)
= O

m
log n
log n

= O

m
log n
log n

= O

m

.
This implies the following theorem:
Minimum Cost Spanning Trees (Kruskal’s Algorithm) 239
Theorem 8.2 Given a graph G with nonnegative weights on its edges
and a source vertex s, Algorithm shortestpath ﬁnds the distance from s
to every other vertex in O(mlog n) time. If the graph is dense, i.e., if m ≥
n
1+
for some > 0, then it can be further improved to run in time O(m/).
8.3 Minimum Cost Spanning Trees (Kruskal’s Algorithm)
Deﬁnition 8.1 Let G = (V, E) be a connected undirected graph with
weights on its edges. A spanning tree (V, T) of G is a subgraph of G that
is a tree. If G is weighted and the sum of the weights of the edges in T is
minimum, then (V, T) is called a minimum cost spanning tree or simply a
minimum spanning tree.
We will assume throughout this section that G is connected. If G is not
connected, then the algorithm can be applied on each connected component
of G. Kruskal’s algorithm works by maintaining a forest consisting of several
spanning trees that are gradually merged until ﬁnally the forest consists of
exactly one tree: a minimum cost spanning tree. The algorithm starts by
sorting the edges in nondecreasing order by weight. Next, starting from the
forest (V, T) consisting of the vertices of the graph and none of its edges,
the following step is repeated until (V, T) is transformed into a tree: Let
(V, T) be the forest constructed so far, and let e ∈ E−T be the current edge
being considered. If adding e to T does not create a cycle, then include e
in T; otherwise discard e. This process will terminate after adding exactly
n −1 edges. The algorithm is summarized below.
1. Sort the edges in G by nondecreasing weight.
2. For each edge in the sorted list, include that edge in the spanning
tree T if it does not form a cycle with the edges currently included
in T; otherwise discard it.
Example 8.3 Consider the weighted graph shown in Fig. 8.4(a). As shown
in Fig. 8.4(b), the ﬁrst edge that is added is (1, 2) since it is of minimum cost.
Next, as shown in Figs. 8.4(c)-(e), edges (1, 3), (4, 6) and then (5, 6) are included
in T in this order. Next, as shown in Fig. 8.4(f), the edge (2, 3) creates a cycle
240 The Greedy Approach
1
2
3
6
4
5
1
2
3
4
9
(h)
1
2
3
6
4
5
1
2
3
4
7
(g)
1
2
3
6
4
5
1
2
3
4
6
(f)
1
2
3
6
4
5
1
2
3
4
(e)
1
2
3
6
4
5
1
2
3
(d)
1
2
3
6
4
5
1
2
(c)
1
2
3
6
4
5
1
2
3
4
6 7
9
11
13
(a)
1
2
3
6
4
5
1
(b)
Fig. 8.4 An Example of Kruskal Algorithm.
and hence is discarded. For the same reason, as shown in Fig. 8.4(g), edge (4, 5)
is also discarded. Finally, edge (3, 4) is included, which results in the minimum
spanning tree (V, T) shown in Fig. 8.4(h).
Implementation of Kruskal’s algorithm
To implement the algorithm eﬃciently, we need some mechanism for testing
whether including an edge creates a cycle. For this purpose, we need to
specify a data structure that represents the forest at each instant of the al-
gorithm and detects cycles dynamically as edges are added to T. A suitable
choice of such data structure is the disjoint sets representation discussed
in Sec. 4.3. In the beginning, each vertex of the graph is represented by a
tree consisting of one vertex. During the execution of the algorithm, each
connected component of the forest is represented by a tree. This method
Minimum Cost Spanning Trees (Kruskal’s Algorithm) 241
is described more formally in Algorithm kruskal. First, the set of edges
is sorted in nondecreasing order by weight. Next n singleton sets are cre-
ated, one for each vertex, and the set of spanning tree edges is initially
empty. The while loop is executed until the minimum cost spanning tree is
constructed.
Algorithm 8.3 kruskal
Input: A weighted connected undirected graph G = (V, E) with n vertices.
Output: The set of edges T of a minimum cost spanning tree for G.
1. Sort the edges in E by nondecreasing weight.
2. for each vertex v ∈ V
3. makeset(¦v¦)
4. end for
5. T = ¦¦
6. while [T[ < n −1
7. Let (x, y) be the next edge in E.
8. if find(x) ,= find(y) then
9. Add (x, y) to T
10. union(x, y)
11. end if
12. end while
Correctness
Lemma 8.2 Algorithm kruskal correctly ﬁnds a minimum cost span-
ning tree in a weighted undirected graph.
Proof. We prove by induction on the size of T that T is a subset of the
set of edges in a minimum cost spanning tree. Initially, T = ¦¦ and the
statement is trivially true. For the induction step, assume before adding
the edge e = (x, y) in Step 7 of the algorithm that T ⊂ T
∗
, where T
∗
is
the set of edges in a minimum cost spanning tree G
∗
= (V, T
∗
). Let X
be the set of vertices in the subtree containing x. Let T

= T ∪ ¦e¦. We
will show that T

is also a subset of the set of edges in a minimum cost
spanning tree. By the induction hypothesis, T ⊂ T
∗
. If T
∗
contains e,
then there is nothing to prove. Otherwise, by Theorem 3.1(c), T
∗
∪ ¦e¦
contains exactly one cycle with e being one of its edges. Since e = (x, y)
connects one vertex in X to another vertex in V −X, T
∗
must also contain
another edge e

would have been added before since it
does not create a cycle with the edges of T
∗
which contains the edges of
T. If we now construct T
∗∗
= T
∗
− ¦e

¦ ∪ ¦e¦, we notice that T

⊂ T
∗∗
.
Moreover, T
∗∗
is the set of edges in a minimum cost spanning tree since e
is of minimum cost among all edges connecting the vertices in X with those
in V −X. ⁄
Time complexity
We analyze the time complexity of the algorithm as follows. Steps 1 and
2 cost O(mlog m) and Θ(n), respectively, where m = [E[. Step 5 costs
Θ(1), and since it is executed at most m times, its total cost is O(m).
Step 7 is executed exactly n −1 times for a total of Θ(n) time. The union
operation is executed n−1 times, and the ﬁnd operation at most 2m times.
By Theorem 4.3, the overall cost of these two operations is O(mlog
∗
n).
Thus, the overall running time of the algorithm is dominated by the sorting
step, i.e., O(mlog m).
Theorem 8.3 Algorithm kruskal ﬁnds a minimum cost spanning tree
in a weighted undirected graph with m edges in O(mlog m) time.
Proof. Lemma 8.2 establishes the correctness of the algorithm and the
time complexity follows from the above discussion. ⁄
Since m can be as large as n(n − 1)/2 = Θ(n
2
), the time complexity
expressed in terms of n is O(n
2
log n). If the graph is planar, then m =
O(n) (see Sec. 3.3.2) and hence the running time of the algorithm becomes
O(nlog n).
8.4 Minimum Cost Spanning Trees (Prim’s Algorithm)
As in the previous section, we will assume throughout this section that G
is connected. If G is not connected, then the algorithm can be applied on
each connected component of G.
This is another algorithm for ﬁnding a minimum cost spanning tree in a
weighted undirected graph that has a totally diﬀerent approach from that
of Algorithm kruskal. Prim’s algorithm for ﬁnding a minimum spanning
tree for an undirected graph is so similar to Dijkstra’s algorithm for the
shortest path problem. The algorithm grows the spanning tree starting
Minimum Cost Spanning Trees (Prim’s Algorithm) 243
from an arbitrary vertex. Let G = (V, E), where for simplicity V is taken
to be the set of integers ¦1, 2, . . . , n¦. The algorithm begins by creating two
sets of vertices: X = ¦1¦ and Y = ¦2, 3, . . . , n¦. It then grows a spanning
tree, one edge at a time. At each step, it ﬁnds an edge (x, y) of minimum
weight, where x ∈ X and y ∈ Y and moves y from Y to X. This edge
is added to the current minimum spanning tree edges in T. This step is
repeated until Y becomes empty. The algorithm is outlined below. It ﬁnds
the set of edges T of a minimum cost spanning tree.
1. T ←¦¦; X ←¦1¦; Y ←V −¦1¦
2. while Y ,= ¦¦
3. Let (x, y) be of minimum weight such that x ∈ X and y ∈ Y .
4. X←X ∪ ¦y¦
5. Y ←Y −¦y¦
6. T ←T ∪ ¦(x, y)¦
7. end while
1
2
3
6
4
5
1
2
3
4
6
7 9
11
13
(e)
1
2
3
6
4
5
1
2
3
4
9
(f)
1
2
3
6
4
5
1
2
3
4
6 7
9
11
13
(c)
1
2
3
6
4
5
1
2
3
4
6 7
9
11
13
(b)
1
2
3
6
4
5
1
2
3
4
6 7
9
11
13
(a)
1
2
3
6
4
5
1
2
3
4
6 7
9
11
13
(d)
Fig. 8.5 An example of Prim’s algorithm.
244 The Greedy Approach
Example 8.4 Consider the graph shown in Fig. 8.5(a). The vertices to the
left of the dashed line belong to X, and those to its right belong to Y . First,
as shown in Fig. 8.5(a), X = ¦1¦ and Y = ¦2, 3, . . . , 6¦. In Fig. 8.5(b), vertex 2
is moved from Y to X since edge (1, 2) has the least cost among all the edges
incident to vertex 1. This is indicated by moving the dashed line so that 1 and 2
are now to its left. As shown in Fig. 8.5(b), the candidate vertices to be moved
from Y to X are 3 and 4. Since edge (1, 3) is of least cost among all edges with
one end in X and one end in Y , 3 is moved from Y to X. Next, from the two
candidate vertices 4 and 5 in Fig. 8.5(c), 4 is moved since the edge (3, 4) has the
least cost. Finally, vertices 6 and then 5 are moved from Y to X as shown in
Fig. 8.5(e). Each time a vertex y is moved from Y to X, its corresponding edge
is included in T, the set of edges of the minimum spanning tree. The resulting
minimum spanning tree is shown in Fig. 8.5(f).
Implementation of Prim’s algorithm
We will assume that the input is represented by adjacency lists. The cost
(i.e. weight) of an edge (x, y), denoted by c[x, y], is stored at the node
for y in the adjacency list corresponding of x. This is exactly the input
representation for Dijkstra’s algorithm shown in Fig. 8.2 (except that here
we are dealing with undirected graphs). The two sets X and Y will be
implemented as boolean vectors X[1..n] and Y [1..n]. Initially, X[1] = 1
and Y [1] = 0, and for all i, 2 ≤ i ≤ n, X[i] = 0 and Y [i] = 1. Thus,
the operation X←X ∪ ¦y¦ is implemented by setting X[y] to 1, and the
operation Y ←Y − ¦y¦ is implemented by setting Y [y] to 0. The set of
tree edges T will be implemented as a linked list, and thus the operation
T ←T ∪ ¦(x, y)¦ simply appends edge (x, y) to T. It is easy to build the
adjacency list representation of the resulting minimum cost spanning tree
from this linked list.
If (x, y) is an edge such that x ∈ X and y ∈ Y , we will call y a bordering
vertex. Bordering vertices are candidates for being moved from Y to X.
Let y be a bordering vertex. Then there is at least one vertex x ∈ X that
is adjacent to y. We deﬁne the neighbor of y, denoted by N[y], to be that
vertex x in X with the property that c[x, y] is minimum among all vertices
adjacent to y in X. We also deﬁne C[y] = c[y, N[y]]. Thus N[y] is the
nearest neighbor to y in X, and C[y] is the cost of the edge connecting y
and N[y]. A detailed description of the algorithm is given as Algorithm
prim.
Initially, we set N[y] to 1 and C[y] = c[1, y] for each vertex y adjacent
Minimum Cost Spanning Trees (Prim’s Algorithm) 245
to 1. For each vertex y that is not adjacent to 1, we set C[y] to ∞. In each
iteration, that vertex y with minimum C[y] is moved from Y to X. After
it has been moved, N[w] and C[w] are updated for each vertex w in Y that
is adjacent to y.
Algorithm 8.4 prim
Input: A weighted connected undirected graph G = (V, E), where
V = ¦1, 2, . . . , n¦.
Output: The set of edges T of a minimum cost spanning tree for G.
1. T ←¦¦; X ←¦1¦; Y ←V −¦1¦
2. for y←2 to n
3. if y adjacent to 1 then
4. N[y] ←1
5. C[y] ←c[1, y]
6. else C[y] ←∞
7. end if
8. end for
9. for j ←1 to n −1 ¦ﬁnd n −1 edges¦
10. Let y ∈ Y be such that C[y] is minimum
11. T ←T ∪ ¦(y, N[y])¦ ¦add edge (y, N[y]) to T¦
12. X←X ∪ ¦y¦ ¦add vertex y to X¦
13. Y ←Y −¦y¦ ¦delete vertex y from Y ¦
14. for each vertex w ∈ Y that is adjacent to y
15. if c[y, w] < C[w] then
16. N[w] ←y
17. C[w] ←c[y, w]
18. end if
19. end for
20. end for
Correctness
Lemma 8.3 Algorithm prim correctly ﬁnds a minimum cost spanning
tree in a connected undirected graph.
Proof. We prove by induction on the size of T that (X, T) is a subtree
of a minimum cost spanning tree. Initially, T = ¦¦ and the statement is
trivially true. Assume the statement is true before adding edge e = (x, y)
in Step 9 of the algorithm, where x ∈ X and y ∈ Y . Let X

= X ∪ ¦y¦
and T

= T ∪¦e¦. We will show that G

= (X

, T

) is also a subset of some
minimum cost spanning tree. First, we show that G

is a tree. Since e is
246 The Greedy Approach
connected to exactly one vertex in X, namely x, and since by the induction
hypothesis (X, T) is a tree, G

is connected and has no cycles, i.e., a tree.
We now show that G

is a subtree of a minimum cost spanning tree. By the
induction hypothesis, T ⊂ T
∗
, where T
∗
is the set of edges in a minimum
spanning tree G
∗
= (V, T
∗
). If T
∗
contains e, then there is nothing to
prove. Otherwise, by Theorem 3.1(c), T
∗
∪ ¦e¦ contains exactly one cycle
with e being one of its edges. Since e = (x, y) connects one vertex in X to
another vertex in Y , T
∗
must also contain another edge e

⊆ T
∗∗
. Moreover, T
∗∗
is the set of edges in a minimum cost
spanning tree since e is of minimum cost among all edges connecting the
vertices in X with those in Y . ⁄
Time complexity
The time complexity of the algorithm is computed as follows. Step 1 costs
Θ(n) time. The for loop in Step 2 requires Θ(n) time. The time taken
by steps 3–6 is O(n). The time taken by Step 8 to search for a vertex y
closest to X is Θ(n) per iteration. This is because the algorithm inspects
each entry in the vector representing the set Y . Since this step is executed
n − 1 times, the overall time taken by Step 8 is Θ(n
2
). Steps 9, 10 and
11 cost Θ(1) time per iteration for a total of Θ(n) time. The for loop in
Step 12 is executed 2m times, where m = [E[. This is because each edge
(y, w) is inspected twice: once when y is moved to X and the other when
w is moved to X. Hence, the overall time required by Step 12 is Θ(m).
The test in Step 13 is executed exactly m times, and Steps 14 and 15 are
executed at most m times. Thus, Steps 12 to 17 cost Θ(m) time. It follows
that the time complexity of the algorithm is Θ(m+n
2
) = Θ(n
2
).
Theorem 8.4 Algorithm prim ﬁnds a minimum cost spanning tree in
a weighted undirected graph with n vertices in Θ(n
2
) time.
Proof. Lemma 8.3 establishes the correctness of the algorithm and the
rest follows from the above discussion. ⁄
8.4.1 A linear time algorithm for dense graphs
Now we improve on Algorithm prim as we did to Algorithm dijkstra in
order to lower its Θ(n
2
) time complexity to O(mlog n) for graphs in which
Minimum Cost Spanning Trees (Prim’s Algorithm) 247
m = o(n
2
). We will also improve it further, so that in the case of dense
graphs it runs in time linear in the number of edges.
As in Algorithm shortestpath, the basic idea is to use the min-heap
data structure (see Sec. 4.2) to maintain the set of bordering vertices so
that the vertex y in Y incident to an edge of lowest cost that is connected
to a vertex in V − Y can be extracted in O(log n) time. The modiﬁed
algorithm is given as Algorithm mst.
Algorithm 8.5 mst
Input: A weighted connected undirected graph G = (V, E), where
V = ¦1, 2, . . . , n¦.
Output: The set of edges T of a minimum cost spanning tree for G.
Assume that we have an empty heap H at the beginning.
1. T ←¦¦; Y ←V −¦1¦
2. for y←2 to n
3. if y is adjacent to 1 then
4. N[y] ←1
5. key(y)←c[1, y]
6. insert(H, y)
7. else key(y)←∞
8. end if
9. end for
10. for j ←1 to n −1 ¦ﬁnd n −1 edges¦
11. y←deletemin(H)
12. T ←T ∪ ¦(y, N[y])¦ ¦add edge (y, N[y]) to T¦
13. Y ←Y −¦y¦ ¦delete vertex y from Y ¦
14. for each vertex w ∈ Y that is adjacent to y
15. if c[y, w] < key(w) then
16. N[w] ←y
17. key(w)←c[y, w]
18. end if
19. if w / ∈ H then insert(H, w)
20. else siftup(H, H
−1
(w))
21. end for
22. end for
The heap H initially contains all vertices adjacent to vertex 1. Each
vertex y ∈ Y is assigned a key which is the cost of the edge connecting y to
1 if it exists; otherwise that key is set to ∞. Each iteration of the for loop
starts by extracting that vertex y with minimum key. The key of each vertex
w in Y adjacent to y is then updated. Next, if w is not in the heap, then
it is inserted; otherwise it is sifted up, if necessary. The function H
−1
(w)
248 The Greedy Approach
returns the position of w in H. This can be implemented by simply having
an array that has for its jth entry the position of vertex j in the heap. As
in Algorithm shortestpath, the running time is dominated by the heap
operations. There are n−1 deletemin operations, n−1 insert operations
and at most m − n + 1 siftup operations. Each one of these operations
takes O(log n) time using binary heaps which results in O(mlog n) time in
total.
If we use a d-heap, the running time is improved as follows. Each
deletemin takes O(d log
d
n) time, and each insert or siftup operation
requires O(log
d
n) time. Thus, the total running time is O(nd log
d
n +
mlog
d
n). If we choose d = 2 +m/n|, the time bound becomes
O(mlog
2+m/n
n). If m ≥ n
1+
for some > 0 that is not too
small, i.e., the graph is dense, then the running time is
O(mlog
2+m/n
n) = O(mlog
2+n

n)
= O

m
log n
log(2 +n

)

= O

m
log n
log n

= O

m

.
This implies the following theorem:
Theorem 8.5 Given a weighted undirected graph G, Algorithm mst
ﬁnds a minimum cost spanning tree in O(mlog n) time. If the graph is
dense, i.e., if m ≥ n
1+
for some > 0, then it can be improved further to
run in time O(m/).
8.5 File Compression
Suppose we are given a ﬁle, which is a string of characters. We wish
to compress the ﬁle as much as possible in such a way that the original
ﬁle can easily be reconstructed. Let the set of characters in the ﬁle be
C = ¦c
1
, c
2
, . . . , c
n
¦. Let also f(c
i
), 1 ≤ i ≤ n, be the frequency of char-
acter c
i
in the ﬁle, i.e., the number of times c
i
appears in the ﬁle. Using
a ﬁxed number of bits to represent each character, called the encoding of
the character, the size of the ﬁle depends only on the number of characters
File Compression 249
in the ﬁle. However, since the frequency of some characters may be much
larger than others, it is reasonable to use variable length encodings. In-
tuitively, those characters with large frequencies should be assigned short
encodings, whereas long encodings may be assigned to those characters with
small frequencies. When the encodings vary in length, we stipulate that
the encoding of one character must not be the preﬁx of the encoding of
another character; such codes are called preﬁx codes. For instance, if we
assign the encodings 10 and 101 to the letters “a” and “b”, there will be
an ambiguity as to whether 10 is the encoding of “a” or is the preﬁx of the
encoding of the letter “b”.
Once the preﬁx constraint is satisﬁed, the decoding becomes unambigu-
ous; the sequence of bits is scanned until an encoding of some character is
found. One way to “parse” a given sequence of bits is to use a full binary
tree, in which each internal node has exactly two branches labeled by 0 and
1. The leaves in this tree correspond to the characters. Each sequence of
0’s and 1’s on a path from the root to a leaf corresponds to a character
encoding. In what follows we describe how to construct a full binary tree
that minimizes the size of the compressed ﬁle.
The algorithm presented is due to Huﬀman. The code constructed by
the algorithm satisﬁes the preﬁx constraint and minimizes the size of the
compressed ﬁle. The algorithm consists of repeating the following procedure
until C consists of only one character. Let c
i
and c
j
be two characters with
minimum frequencies. Create a new node c whose frequency is the sum
of the frequencies of c
i
and c
j
, and make c
i
and c
j
the children of c. Let
C = C −¦c
i
, c
j
¦ ∪ ¦c¦.
4
d c
10
/
7
o
20
11
21
c
18
38
59
1
2
3
4
0
1
0
0
0
1
1
1
Fig. 8.6 An example of a Huﬀman tree.
250 The Greedy Approach
Example 8.5 Consider ﬁnding a preﬁx code for a ﬁle that consists of the
letters a, b, c, d, and e. See Fig. 8.6. Suppose that these letters appear in the ﬁle
with the following frequencies.
f(a) = 20, f(b) = 7, f(c) = 10, f(d) = 4 and f(e) = 18.
Each leaf node is labeled with its corresponding character and its frequency of
occurrence in the ﬁle. Each internal node is labeled with the sum of the weights
of the leaves in its subtree and the order of its creation. For instance, the ﬁrst
internal node created has a sum of 11 and it is labeled with 1. From the binary
tree, the encodings for a, b, c, d, and e are, respectively, 01, 110, 10, 111 and 00.
Suppose each character was represented by three binary digits before compression.
Then, the original size of the ﬁle is 3(20+7+10+4+18) = 177 bits. The size of the
compressed ﬁle using the above code becomes 220+37+210+34+218 =
129 bits, a saving of about 27%.
The algorithm
Since the main operations required to construct a Huﬀman tree are insertion
and deletion of characters with minimum frequency, a min-heap is a good
candidate data structure that supports these operations. The algorithm
builds a tree by adding n − 1 internal vertices one at a time; its leaves
correspond to the input characters. Initially and during its execution, the
algorithm maintains a forest of trees. After adding n − 1 internal nodes,
the forest is transformed into a tree: the Huﬀman tree. Below, we give a
more precise description of an algorithm that constructs a full binary tree
corresponding to a Huﬀman code of an input string of characters together
with their frequencies.
Time complexity
The time complexity of the algorithm is computed as follows. The time
needed to insert all characters into the heap is Θ(n) (Theorem 4.1). The
time required to delete two elements from the heap and add a new element
is O(log n). Since this is repeated n−1 times, the overall time taken by the
for loop is O(nlog n). It follows that the time complexity of the algorithm
is O(nlog n).
Exercises 251
Algorithm 8.6 huffman
Input: A set C = ¦c
1
, c
2
, . . . , c
n
¦ of n characters and their frequencies
¦f(c
1
), f(c
2
), . . . , f(c
n
)¦.
Output: A Huﬀman tree (V, T) for C.
1. Insert all characters into a min-heap H according to their frequencies.
2. V ←C; T = ¦¦
3. for j ←1 to n −1
4. c←deletemin(H)
5. c

children of v in T¦
10. end while
8.6 Exercises
8.1. Is Algorithm linearsearch described in Sec. 1.3 a greedy algorithm?
Explain.
8.2. Is Algorithm majority described in Sec. 5.7 a greedy algorithm? Ex-
plain.
8.3. This exercise is about the money change problem stated in Exercise 7.30.
Consider a currency system that has the following coins and their values:
dollar (100 cents), quarter (25 cents), dime (10 cents), nickel (5 cents)
and 1-cent coins. (A unit-value coin is always required). Suppose we
want to give a change of value n cents in such a way that the total
number of coins n is minimized. Give a greedy algorithm to solve this
problem.
8.4. Give a counterexample to show that the greedy algorithm obtained in
Exercise 8.3 does not always work if we instead use coins of values 1
cent, 5 cents, 7 cents and 11 cents. Note that in this case dynamic
programming can be used to ﬁnd the minimum number of coins. (See
Exercises 7.30 and 7.31).
8.5. Suppose in the money change problem of Exercise 8.3 the coin values
are: 1, 2, 4, 8, 16, . . . , 2
k
, for some positive integer k. Give an O(log n)
algorithm to solve the problem if the value to be paid is n < 2
k+1
.
8.6. For what denominations ¦v
1
, v
2
, . . . , v
k
¦, k ≥ 2, does the greedy algo-
rithm for the money change problem stated in Exercise 7.30 always give
the minimum number of coins? Prove your answer.
8.7. Let G = (V, E) be an undirected graph. A vertex cover for G is a subset
252 The Greedy Approach
S ⊆ V such that every edge in E is incident to at least one vertex in S.
Consider the following algorithm for ﬁnding a vertex cover for G. First,
order the vertices in V by decreasing order of degree. Next execute the
following step until all edges are covered. Pick a vertex of highest degree
that is incident to at least one edge in the remaining graph, add it to the
cover, and delete all edges incident to that vertex. Show that this greedy
approach does not always result in a vertex cover of minimum size.
8.8. Let G = (V, E) be an undirected graph. A clique C in G is a subgraph
of G that is a complete graph by itself. A clique C is maximum if there
is no other clique C

in G such that the size of C

is greater than the size
of C. Consider the following method that attempts to ﬁnd a maximum
clique in G. Initially, let C = G. Repeat the following step until C is
a clique. Delete from C a vertex that is not connected to every other
vertex in C. Show that this greedy approach does not always result in a
maximum clique.
8.9. Let G = (V, E) be an undirected graph. A coloring of G is an assignment
of colors to the vertices in V such that no two adjacent vertices have the
same color. The coloring problem is to determine the minimum number
of colors needed to color G. Consider the following greedy method that
attempts to solve the coloring problem. Let the colors be 1, 2, 3, . . ..
First, color as many vertices as possible using color 1. Next, color as
many vertices as possible using color 2, and so on. Show that this greedy
approach does not always color the graph using the minimum number of
colors.
8.10. Let A
1
, A
2
, . . . , A
m
be m arrays of integers each sorted in nondecreas-
ing order. Each array A
j
is of size n
j
. Suppose we want to merge all
arrays into one array A using an algorithm similar to Algorithm merge
described in Sec. 1.4. Give a greedy strategy for the order in which these
arrays should be merged so that the overall number of comparisons is
minimized. For example, if m = 3, we may merge A
1
with A
2
to obtain
A
4
and then merge A
3
with A
4
to obtain A. Another alternative is to
merge A
2
with A
3
to obtain A
4
and then merge A
1
with A
4
to obtain
A. Yet another alternative is to merge A
1
with A
3
to obtain A
4
and
then merge A
2
with A
4
to obtain A. (Hint: Give an algorithm similar
to Algorithm huffman).
8.11. Analyze the time complexity of the algorithm in Exercise 8.10.
8.12. Consider the following greedy algorithm which attempts to ﬁnd the dis-
tance from vertex s to vertex t in a directed graph G with positive lengths
on its edges. Starting from vertex s, go to the nearest vertex, say x. From
vertex x, go to the nearest vertex, say y. Continue in this manner until
you arrive at vertex t. Give a graph with the fewest number of vertices
to show that this heuristic does not always produce the distance from s
Exercises 253
to t. (Recall that the distance from vertex u to vertex v is the length of
a shortest path from u to v).
8.13. Apply Algorithm dijkstra on the directed graph shown in Fig. 8.7.
Assume that vertex 1 is the start vertex.
1
2
3
6
4
5
12
15
4
9
13
4
3
5
2
Fig. 8.7 Directed graph.
8.14. Is Algorithm dijkstra optimal? Explain.
8.15. What are the merits and demerits of using the adjacency matrix represen-
tation instead of the adjacency lists in the input to Algorithm dijkstra?
8.16. Modify Algorithmdijkstra so that it ﬁnds the shortest paths in addition
to their lengths.
8.17. Prove that the subgraph deﬁned by the paths obtained from the modiﬁed
shortest path algorithm as described in Exercise 8.16 is a tree. This tree
is called the shortest path tree.
8.18. Can a directed graph have two distinct shortest path trees (see Exer-
cise 8.17)? Prove your answer.
8.19. Give an example of a directed graph to show that Algorithm dijkstra
does not always work if some of the edges have negative weights.
8.20. Show that the proof of correctness of Algorithm dijkstra (Lemma 8.1)
does not work if some of the edges in the input graph have negative
weights.
8.21. Let G = (V, E) be a directed graph such that removing the directions
from its edges results in a planar graph. What is the running time of
Algorithm shortestpath when applied to G? Compare that to the
running time when using Algorithm dijkstra.
8.22. Let G = (V, E) be a directed graph such that m = O(n
1.2
), where
n = [V [ and m = [E[. What changes should be made to Algorithm
shortestpath so that it will run in time O(m)?
8.23. Show the result of applying Algorithm kruskal to ﬁnd a minimum cost
spanning tree for the undirected graph shown in Fig. 8.8.
254 The Greedy Approach
1 3 5
6 4 2
1
2
3
3
6
7
6
4
9
7
2
Fig. 8.8 An undirected graph.
8.24. Show the result of applying Algorithm prim to ﬁnd a minimum cost
spanning tree for the undirected graph shown in Fig. 8.8.
8.25. Let G = (V, E) be an undirected graph such that m = O(n
1.99
), where
n = [V [ and m = [E[. Suppose you want to ﬁnd a minimum cost
spanning tree for G. Which algorithm would you choose: Algorithm
prim or Algorithm kruskal? Explain.
8.26. Let e be an edge of minimum weight in an undirected graph G. Show
that e belongs to some minimum cost spanning tree of G.
8.27. Does Algorithm prim work correctly if the graph has negative weights?
Prove your answer.
8.28. Let G be an undirected weighted graph such that no two edges have the
same weight. Prove that G has a unique minimum cost spanning tree.
8.29. What is the number of spanning trees of a complete undirected graph G
with n vertices? For example, the number of spanning trees of K
3
, the
complete graph with three vertices, is 3.
8.30. Let G be a directed weighted graph such that no two edges have the
same weight. Let T be a shortest path tree for G (see Exercise 8.17).
Let G

be the undirected graph obtained by removing the directions from
the edges of G. Let T

be a minimum spanning tree for G

. Prove or
disprove that T = T

.
8.31. Use Algorithm huffman to ﬁnd an optimal code for the characters a, b,
c, d, e and f whose frequencies in a given text are respectively 7, 5, 3, 2,
12, 9.
8.32. Prove that the graph obtained in Algorithm huffman is a tree.
8.33. Algorithm huffman constructs the code tree in a bottom-up fashion. Is
it a dynamic programming algorithm?
8.34. Let B = ¦b
1
, b
2
, . . . , b
n
¦ and W = ¦w
1
, w
2
, . . . , w
n
¦ be two sets of black
and white points in the plane. Each point is represented by the pair
(x, y) of x and y coordinates. A black point b
i
= (x
i
, y
i
) dominates
a white point w
j
= (x
j
, y
j
) if and only if x
i
≥ x
j
and y
i
≥ y
j
. A
matching between a black point b
i
and a white point w
j
is possible if
Bibliographic notes 255
b
i
dominates w
j
. A matching M = ¦(b
i
1
, w
j
1
), (b
i
2
, w
j
2
), . . . , (b
i
k
, w
j
k
)¦
between the black and white points is maximum if k, the number of
matched pairs in M, is maximum. Design a greedy algorithm to ﬁnd a
maximum matching in O(nlog n) time. (Hint: Sort the black points in
increasing x-coordinates and use a heap for the white points).
8.7 Bibliographic notes
The greedy graph algorithms are discussed in most books on algorithms
(see the bibliographic notes of Chapter 1).
Algorithm dijkstra for the single source shortest path problem is from
Dijkstra (1959). The implementation using a heap is due to Johnson (1977);
see also Tarjan (1983). The best known asymptotic running time for this
problem is O(m+nlog n), which is due to Fredman and Tarjan (1987).
Graham and Hell (1985) discuss the long history of the minimum cost
spanning tree problem, which has been extensively studied. Algorithm
kruskal comes from Kruskal (1956). Algorithm prim is due to Prim
(1957). The improvement using heaps can be found in Johnson (1975).
More sophisticated algorithms can be found in Yao (1975), Cheriton and
Tarjan (1976) and Tarjan (1983). Algorithm huffman for ﬁle compression
is due to Huﬀman (1952) (see also Knuth (1968)).
256
Chapter 9
Graph Traversal
9.1 Introduction
In some graph algorithms such as those for ﬁnding shortest paths or min-
imum spanning trees, the vertices and edges are visited in an order that
is imposed by their respective algorithms. However, in some other algo-
rithms, the order of visiting the vertices is unimportant; what is important
is that the vertices are visited in a systematic order, regardless of the in-
put graph. In this chapter, we discuss two methods of graph traversal:
depth-ﬁrst search and breadth-ﬁrst search.
9.2 Depth-First Search
Depth-ﬁrst search is a powerful traversal method that aids in the solution
of many problems involving graphs. It is essentially a generalization of the
preorder traversal of rooted trees (see Sec. 3.5.1). Let G = (V, E) be a
directed or undirected graph. A depth-ﬁrst search traversal of G works as
follows. First, all vertices are marked unvisited. Next, a starting vertex
is selected, say v ∈ V , and marked visited. Let w be any vertex that is
adjacent to v. We mark w as visited and advance to another vertex, say
x, that is adjacent to w and is marked unvisited. Again, we mark x as
visited and advance to another vertex that is adjacent to x and is marked
unvisited. This process of selecting an unvisited vertex adjacent to the
current vertex continues as deep as possible until we ﬁnd a vertex y whose
adjacent vertices have all been marked visited. At this point, we back up to
the most recently visited vertex, say z, and visit an unvisited vertex that
257
258 Graph Traversal
is adjacent to z, if any. Continuing this way, we ﬁnally return back to the
starting vertex v. This method of traversal has been given the name depth-
ﬁrst search, as it continues the search in the forward (deeper) direction.
The algorithm for such a traversal can be written using recursion as shown
in Algorithm dfs or a stack (see Exercise 9.5).
Algorithm 9.1 dfs
Input: A (directed or undirected) graph G = (V, E).
Output: Preordering and postordering of the vertices in the corresponding
depth-ﬁrst search tree.
1. predfn ←0; postdfn ←0
2. for each vertex v ∈ V
3. mark v unvisited
4. end for
5. for each vertex v ∈ V
6. if v is marked unvisited then dfs(v)
7. end for
Procedure dfs(v)
1. mark v visited
2. predfn ←predfn + 1
3. for each edge (v, w) ∈ E
4. if w is marked unvisited then dfs(w)
5. end for
6. postdfn ←postdfn + 1
The algorithm starts by marking all vertices unvisited. It also initializes
two counters predfn and postdfn to zero. These two counters are not part
of the traversal ; their importance will be apparent when we later make
use of depth-ﬁrst search to solve some problems. The algorithm then calls
Procedure dfs for each unvisited vertex in V . This is because not all the
vertices may be reachable from the start vertex. Starting from some vertex
v ∈ V , Procedure dfs performs the search on G by visiting v, marking v
visited and then recursively visiting its adjacent vertices. When the search
is complete, if all vertices are reachable from the start vertex, a spanning
tree called the depth-ﬁrst search spanning tree is constructed whose edges
are those inspected in the forward direction, i.e., when exploring unvisited
vertices. In other words, let (v, w) be an edge such that w is marked
unvisited and suppose the procedure was invoked by the call dfs(v). Then,
in this case, that edge will be part of the depth-ﬁrst search spanning tree.
Depth-First Search 259
If not all the vertices are reachable from the start vertex, then the search
results in a forest of spanning trees instead. After the search is complete,
each vertex is labeled with predfn and postdfn numbers. These two labels
impose preorder and postorder numbering on the vertices in the spanning
tree (or forest) generated by the depth-ﬁrst search traversal. They give the
order in which visiting a vertex starts and ends. In the following, we say
that edge (v, w) is being explored to mean that within the call dfs(v), the
procedure is inspecting the edge (v, w) to test whether w has been visited
before or not. The edges of the graph are classiﬁed diﬀerently according to
whether the graph is directed or undirected.
The case of undirected graphs
Let G = (V, E) be an undirected graph. As a result of the traversal, the
edges of G are classiﬁed into the following two types:
• Tree edges: edges in the depth-ﬁrst search tree. An edge (v, w) is
a tree edge if w was ﬁrst visited when exploring the edge (v, w).
• Back edges: All other edges.
back edges tree edges
(a)
o
;
o
/ c
c
d
/
,
j
1,10
o
d
c
2,9
3,3
4,2
5,1
10,4
;
j
,
6,8
7,7
8,6
9,5
(b)
/
o
c
/
Fig. 9.1 An example of depth-ﬁrst search traversal of an undirected graph.
Example 9.1 Figure 9.1(b) illustrates the action of depth-ﬁrst search traver-
sal on the undirected graph shown in Fig. 9.1(a). Vertex a has been selected as
260 Graph Traversal
the start vertex. The depth-ﬁrst search tree is shown in Fig. 9.1(b) with solid
lines. Dotted lines represent back edges. Each vertex in the depth-ﬁrst search
tree is labeled with two numbers: predfn and postdfn. Note that since vertex e
has postdfn = 1, it is the ﬁrst vertex whose depth-ﬁrst search is complete. Note
also that since the graph is connected, the start vertex is labeled with predfn = 1
and postdfn = 10, the number of vertices in the graph.
The case of directed graphs
Depth-ﬁrst search in directed graphs results in one or more (directed) span-
ning trees whose number depends on the start vertex. If v is the start vertex,
depth-ﬁrst search generates a tree consisting of all vertices reachable from
v. If not all vertices are included in that tree, the search resumes from an-
other unvisited vertex, say w, and a tree consisting of all unvisited vertices
that are reachable from w is constructed. This process continues until all
vertices have been visited. In depth-ﬁrst search traversal of directed graphs,
however, the edges of G are classiﬁed into four types:
• Tree edges: edges in the depth-ﬁrst search tree. An edge (v, w) is
a tree edge if w was ﬁrst visited when exploring the edge (v, w).
• Back edges: edges of the form (v, w) such that w is an ancestor of
v in the depth-ﬁrst search tree (constructed so far) and vertex w
was marked visited when (v, w) was explored.
• Forward edges: edges of the form (v, w) such that w is a descendant
of v in the depth-ﬁrst search tree (constructed so far) and vertex w
was marked visited when (v, w) was explored.
• Cross edges: All other edges.
Example 9.2 Figure 9.2(b) illustrates the action of depth-ﬁrst search traver-
sal on the directed graph shown in Fig. 9.2(a). Starting at vertex a, the vertices
a, b, e and f are visited in this order. When Procedure dfs is initiated again at
vertex c, vertex d is visited and the traversal is complete after b is visited from
c. We notice that the edge (e, a) is a back edge since e is a descendant of a in
the depth-ﬁrst search tree, and (e, a) is not a tree edge. On the other hand, edge
(a, f) is a forward edge since a is an ancestor of f in the depth-ﬁrst search tree,
and (a, f) is not a tree edge. Since neither e nor f is an ancestor of the other in
the depth-ﬁrst search tree, edge (f, e) is a cross edge. The two edges (c, b) and
(d, e) are, obviously, cross edges; each edge connects two vertices in two diﬀerent
trees. Note that had we chosen to visit vertex f immediately after a instead of
Depth-First Search 261
visiting vertex b, both edges (a, b) and (a, f) would have been tree edges. In this
case, the result of the depth-ﬁrst search traversal is shown in Fig. 9.2(c). Thus
the type of an edge depends on the order in which the vertices are visited.
2,2
3,1
(c)
1,4
5,6
6,5
4,3
d c
;
o
c
/
2,3
4,2
(b)
1,4
5,6
6,5
3,1
;
e
/
o
c
d
(a)
tree edges
back edges
o /
c
d c
;
forward edges
cross edges
Fig. 9.2 An example of depth-ﬁrst search traversal of a directed graph.
9.2.1 Time complexity of depth-ﬁrst search
Now we analyze the time complexity of Algorithm dfs when applied to
a graph G with n vertices and m edges. The number of procedure calls
is exactly n since once the procedure is invoked on vertex v, it will be
marked visited and hence no more calls on v will take place. The cost of a
procedure call if we exclude the for loop is Θ(1). It follows that the overall
cost of procedure calls excluding the for loop is Θ(n). Steps 1 and 2 of the
algorithm cost Θ(1) and Θ(n) time, respectively. The cost of Step 3 of the
algorithm to test whether a vertex is marked is Θ(n). Now, it remains to
ﬁnd the cost of the for loop in Procedure dfs. The number of times this
step is executed to test whether a vertex w is marked unvisited is equal
262 Graph Traversal
to the number of vertices adjacent to vertex v. Hence, the total number
of times this step is executed is equal to the number of edges in the case
of directed graphs and twice the number of edges in the case of undirected
graphs. Consequently, the cost of this step is Θ(m) in both directed and
undirected graphs. It follows that the running time of the algorithm is
Θ(m+n). If the graph is connected or m ≥ n, then the running time is
simply Θ(m). It should be emphasized, however, that the graph is assumed
to be represented by adjacency lists. The time complexity of Algorithm dfs
when the graph is represented by an adjacency matrix is left as an exercise
(Exercise 9.6).
9.3 Applications of Depth-First Search
Depth-ﬁrst search is used quite often in graph and geometric algorithms.
It is a powerful tool and has numerous applications. In this section, we list
some of its important applications.
9.3.1 Graph acyclicity
Let G = (V, E) be a directed or undirected graph with n vertices and m
edges. To test whether G has at least one cycle, we apply depth-ﬁrst search
on G. If a back edge is detected during the search, then G is cyclic; other-
wise G is acyclic. Note that G may be disconnected. If G is a connected
undirected graph, then there is no need for graph traversal, as G is acyclic
if and only if it is a tree, if and only if m = n −1 (Theorem 3.1).
9.3.2 Topological sorting
Given a directed acyclic graph (dag for short) G = (V, E), the problem
of topological sorting is to ﬁnd a linear ordering of its vertices in such a
way that if (v, w) ∈ E, then v appears before w in the ordering. For
example, one possible topological sorting of the vertices in the dag shown
in Fig. 9.3(a) is a, b, d, c, e, f, g. We will assume that the dag has only one
vertex, say s, of indegree 0. If not, we may simply add a new vertex s and
edges from s to all vertices of indegree 0 (see Fig. 9.3(b)).
Next, we simply carry out a depth-ﬁrst search on G starting at vertex s.
When the traversal is complete, the values of the counter postdfn deﬁne
a reverse topological ordering of the vertices in the dag. Thus, to obtain
Applications of Depth-First Search 263
;
c c
d
o
/
o
·
;
c c
d
o
/
o
(a) (b)
Fig. 9.3 Illustration of topological sorting.
the ordering, we may add an output step to Algorithm dfs just after the
counter postdfn is incremented. The resulting output is reversed to obtain
the desired topological ordering.
9.3.3 Finding articulation points in a graph
A vertex v in an undirected graph G with more than two vertices is called
an articulation point if there exist two vertices u and w diﬀerent from v
such that any path between u and w must pass through v. Thus, if G is
connected, the removal of v and its incident edges will result in a discon-
nected subgraph of G. A graph is called biconnected if it is connected and
has no articulation points. To ﬁnd the set of articulation points, we per-
form a depth-ﬁrst search traversal on G. During the traversal, we maintain
two labels with each vertex v ∈ V : α[v] and β[v]. α[v] is simply predfn
in the depth-ﬁrst search algorithm, which is incremented at each call to
the depth-ﬁrst search procedure. β[v] is initialized to α[v], but may change
later on during the traversal. For each vertex v visited, we let β[v] be the
minimum of the following:
• α[v].
• α[u] for each vertex u such that (v, u) is a back edge.
• β[w] for each edge (v, w) in the depth-ﬁrst search tree.
The articulation points are determined as follows
• The root is an articulation point if and only if it has two or more
children in the depth-ﬁrst search tree.
• A vertex v other than the root is an articulation point if and only
if v has a child w with β[w] ≥ α[v].
264 Graph Traversal
The formal algorithm for ﬁnding the articulation points is given as Al-
gorithm articpoints.
Algorithm 9.2 articpoints
Input: A connected undirected graph G = (V, E).
Output: Array A[1..count] containing the articulation points of G, if any.
1. Let s be the start vertex.
2. for each vertex v ∈ V
3. mark v unvisited
4. end for
5. predfn ←0; count ←0; rootdegree ←0
6. dfs(s)
Procedure dfs(v)
1. mark v visited; artpoint ←false ; predfn ←predfn + 1
2. α[v] ←predfn; β[v] ←predfn ¦Initialize α[v] and β[v]¦
3. for each edge (v, w) ∈ E
4. if (v, w) is a tree edge then
5. dfs(w)
6. if v = s then
7. rootdegree ←rootdegree + 1
8. if rootdegree = 2 then artpoint ←true
9. else
10. β[v] ← min¦β[v], β[w]¦
11. if β[w] ≥ α[v] then artpoint ←true
12. end if
13. else if (v, w) is a back edge then β[v] ← min¦β[v], α[w]¦
14. else do nothing ¦w is the parent of v¦
15. end if
16. end for
17. if artpoint then
18. count ←count + 1
19. A[count] ←v
20. end if
First, the algorithm performs the necessary initializations. In particular,
count is the number of articulation points, and rootdegree is the degree of
the root of the depth-ﬁrst search tree. This is needed to decide later whether
the root is an articulation point as mentioned above. Next the depth-ﬁrst
search commences starting at the root. For each vertex v visited, α[v] and
β[v] are initialized to predfn. When the search backs up from some vertex
w to v, two actions take place. First, β[v] is set to β[w] if β[w] is found
to be smaller then β[v]. Second, if β[w] ≥ α[v], then this is an indication
Applications of Depth-First Search 265
that v is an articulation point. This is because any path from w to an
ancestor of v must pass through v. This is illustrated in Fig. 9.4 in which
any path from the subtree rooted at w to u must include v, and hence v
is an articulation point. The subtree rooted at w contains one or more
connected components. In this ﬁgure, the root u is an articulation point
since its degree is greater than 1.
. n
·
j
&
a
Fig. 9.4 Illustration of the algorithm for ﬁnding the articulation points.
Example 9.3 We illustrate the action of Algorithm articpoints by ﬁnding
the articulation points of the graph shown in Fig. 9.1(a). See Fig. 9.5. Each
vertex v in the depth-ﬁrst search tree is labeled with α[v] and β[v]. The depth-
ﬁrst search starts at vertex a and proceeds to vertex e. A back edge (e, c) is
discovered, and hence β[e] is assigned the value α[c] = 3. Now, when the search
backs up to vertex d, β[d] is assigned β[e] = 3. Similarly, when the search backs
up to vertex c, its label β[c] is assigned the value β[d] = 3. Now, since β[d] ≥ α[c],
vertex c is marked as an articulation point. When the search backs up to b, it is
also found that β[c] ≥ α[b], and hence b is also marked as an articulation point.
At vertex b, the search branches to a new vertex f and proceeds, as illustrated
in the ﬁgure, until it reaches vertex j. The back edge (j, h) is detected and hence
β[j] is set to α[h] = 8. Now, as described before, the search backs up to i and
then h and sets β[i] and β[h] to β[j] = 8. Again, since β[i] ≥ α[h], vertex h is
marked as an articulation point. For the same reason, vertex g is marked as an
266 Graph Traversal
o
;
o
/
back edges tree edges
c
c
d
/
,
j
d
c
1,1
¬,1
·,·
i,·
j,·
1o,S
/
c
;
j
,
6,1
¬,1
S,S
o,S
/
o
o
Fig. 9.5 An example of ﬁnding the articulation points in a graph.
articulation point. At vertex g, the back edge (g, a) is detected, and hence β[g] is
set to α[a] = 1. Finally, β[f] and then β[b] are set to 1 and the search terminates
at the start vertex. The root a is not an articulation point since it has only one
child in the depth-ﬁrst search tree.
9.3.4 Strongly connected components
Given a directed graph G = (V, E), a strongly connected component in G
is a maximal set of vertices in which there is a path between each pair of
vertices. Algorithm strongconnectcomp below uses depth-ﬁrst search
in order to identify all the strongly connected components in a directed
graph.
Example 9.4 Consider the directed graph G shown in Fig. 9.2(a). Applying
depth-ﬁrst search on this directed graph results in the forest shown in Fig. 9.2(b).
Also shown in the ﬁgure is the postordering of the vertices, which is e, f, b, a, d, c.
If we reverse the direction of the edges in G, we obtain G

by reversing the direction of edges in G.
3. Perform a depth-ﬁrst search on G

starting from the vertex with
highest postdfn number. If the depth-ﬁrst search does not reach
all vertices, start the next depth-ﬁrst search from the vertex with
highest postdfn number among the remaining vertices.
4. Each tree in the resulting forest corresponds to a strongly connected
component.
at vertex a yields the tree whose vertices are a, b, e and f. The resulting forest is
shown in Fig. 9.6(b). Each tree in the forest corresponds to a strongly connected
component. Thus, G contains three strongly connected components.
(a)
back edges
cross edges
o / c
d c ;
tree edges
(b)
;
c
/
o
c
d
Fig. 9.6 Finding strongly connected components.
9.4 Breadth-First Search
Unlike depth-ﬁrst search, in breadth-ﬁrst search when we visit a vertex
v we next visit all vertices adjacent to v. The resulting tree is called a
breadth-ﬁrst search tree. This method of traversal can be implemented by a
queue to store unexamined vertices. Algorithm bfs for breadth-ﬁrst search
can be applied to directed and undirected graphs. Initially, all vertices are
marked unvisited. The counter bfn, which is initialized to zero, represents
268 Graph Traversal
the order in which the vertices are removed from the queue. In the case
of undirected graphs, an edge is either a tree edge or a cross edge. If the
graph is directed, an edge is either a tree edge, a back edge or a cross edge;
there are no forward edges.
Algorithm 9.4 bfs
Input: A directed or undirected graph G = (V, E).
Output: Numbering of the vertices in breadth-ﬁrst search order.
1. bfn ←0
2. for each vertex v ∈ V
3. mark v unvisited
4. end for
5. for each vertex v ∈ V
6. if v is marked unvisited then bfs(v)
7. end for
Procedure bfs(v)
1. Q← ¦v¦
2. mark v visited
3. while Q ,= ¦¦
4. v←Pop(Q)
5. bfn ←bfn + 1
6. for each edge (v, w) ∈ E
7. if w is marked unvisited then
8. Push(w, Q)
9. mark w visited
10. end if
11. end for
12. end while
Example 9.5 Figure 9.7 illustrates the action of breadth-ﬁrst search traversal
when applied on the graph shown in Fig. 9.1(a) starting from vertex a. After
popping oﬀ vertex a, vertices b and g are pushed into the queue and marked
visited. Next, vertex b is removed from the queue, and its adjacent vertices that
have not yet been visited, namely c and f, are pushed into the queue and marked
visited. This process of pushing vertices into the queue and removing them later
on is continued until vertex j is ﬁnally removed from the queue. At this point,
the queue becomes empty and the breadth-ﬁrst search traversal is complete. In
the ﬁgure, each vertex is labeled with its bfn number, the order in which that
vertex was removed from the queue. Notice that the edges in the ﬁgure are either
Applications of Breadth-First Search 269
cross edges tree edges
o
;
/
c
, j
o
d c
/
1
2 3
4 5 6
7 8 9 10
Fig. 9.7 An example of breadth-ﬁrst search traversal of an undirected graph.
tree edges or cross edges.
Time complexity
The time complexity of breadth-ﬁrst search when applied to a graph (di-
rected or undirected) with n vertices and m edges is the same as that of
depth-ﬁrst search, i.e., Θ(n +m). If the graph is connected or m ≥ n, then
the time complexity is simply Θ(m).
9.5 Applications of Breadth-First Search
We close this chapter with an application of breadth-ﬁrst search that is
important in graph and network algorithms. Let G = (V, E) be a connected
undirected graph and s a vertex in V . When Algorithm bfs is applied to
G starting at s, the resulting breadth-ﬁrst search tree is such that the path
from s to any other vertex has the least number of edges. Thus, suppose we
want to ﬁnd the distance from s to every other vertex, where the distance
from s to a vertex v is deﬁned to be the least number of edges in any path
from s to v. This can easily be done by labeling each vertex with its distance
prior to pushing it into the queue. Thus, the start vertex will be labeled 0,
its adjacent vertices with 1, and so on. Clearly, the label of each vertex is its
shortest distance from the start vertex. For instance, in Fig. 9.7, vertex a
will be labeled 0, vertices b and g will be labeled 1, vertices c, f and h will
be labeled 2, and ﬁnally vertices d, e, i and j will be labeled 3. Note that
this vertex numbering is not the same as the breadth-ﬁrst numbering in
the algorithm. The minor changes to the breadth-ﬁrst search algorithm are
270 Graph Traversal
left as an exercise (Exercise 9.26).
9.6 Exercises
9.1. Show the result of running depth-ﬁrst search on the undirected graph
shown in Fig. 9.8(a) starting at vertex a. Give the classiﬁcation of edges
as tree edges or back edges.
9.2. Show the result of running depth-ﬁrst search on the directed graph shown
in Fig. 9.8(b) starting at vertex a. Give the classiﬁcation of edges as tree
edge, back edges, forward edges or cross edges.
;
c
d
o
c
/
(b)
;
c
d
o
c
/
(a)
Fig. 9.8 Undirected and directed graphs.
9.3. Show the result of running depth-ﬁrst search on the undirected graph of
Fig. 9.9 starting at vertex f. Give the classiﬁcation of edges.
j
,
/
o
; c c
d
/
o
Fig. 9.9 An undirected graph.
9.4. Show the result of running depth-ﬁrst search on the directed graph of
Fig. 9.10 starting at vertex e. Give the classiﬁcation of edges.
9.5. Give an iterative version of Algorithm dfs that uses a stack to store
unvisited vertices.
Exercises 271
o
; d
c c
/
/
o
Fig. 9.10 A directed graph.
9.6. What will be the time complexity of the depth-ﬁrst search algorithm if
the input graph is represented by an adjacency matrix (see Sec. 3.3.1 for
graph representation).
9.7. Show that when depth-ﬁrst search is applied to an undirected graph G,
the edges of G will be classiﬁed as either tree edges or back edges. That
is, there are no forward edges or cross edges.
9.8. Suppose that Algorithm dfs is applied to an undirected graph G. Give
an algorithm that classiﬁes the edges of G as either tree edges or back
edges.
9.9. Suppose that Algorithm dfs is applied to a directed graph G. Give an
algorithm that classiﬁes the edges of G as either tree edges, back edges,
forward edges or cross edges.
9.10. Give an algorithm that counts the number of connected components in
an undirected graph using depth-ﬁrst search or breadth-ﬁrst search.
9.11. Given an undirected graph G, design an algorithm to list the vertices in
each connected component of G separately.
9.12. Give an O(n) time algorithm to determine whether a connected undi-
rected graph with n vertices contains a cycle.
9.13. Apply the articulation points algorithm to obtain the articulation points
of the undirected graph shown in Fig. 9.9.
9.14. Let T be the depth-ﬁrst search tree resulting from a depth-ﬁrst search
traversal on a connected undirected graph. Show that the root of T is
an articulation point if and only if it has two or more children. (See
Sec. 9.3.3)
9.15. Let T be the depth-ﬁrst search tree resulting from a depth-ﬁrst search
traversal on a connected undirected graph. Show that a vertex v other
than the root is an articulation point if and only if v has a child w with
β[w] ≥ α[v]. (See Sec. 9.3.3)
9.16. Apply the strongly connected components algorithm on the directed
graph shown in Fig. 9.10.
9.17. Show that in the strongly connected components algorithm, any choice
272 Graph Traversal
of the ﬁrst vertex to carry out the depth-ﬁrst search traversal leads to
the same solution.
9.18. An edge of a connected undirected graph Gis called a bridge if its deletion
disconnects G. Modify the algorithm for ﬁnding articulation points so
that it detects bridges instead of articulation points.
9.19. Show the result of running breadth-ﬁrst search on the undirected graph
shown in Fig. 9.8(a) starting at vertex a.
9.20. Show the result of running breadth-ﬁrst search on the directed graph
shown in Fig. 9.8(b) starting at vertex a.
9.21. Show the result of running breadth-ﬁrst search on the undirected graph
of Fig. 9.1 starting at vertex a.
9.22. Show the result of running breadth-ﬁrst search on the undirected graph
of Fig. 9.9 starting at vertex f.
9.23. Show the result of running breadth-ﬁrst search on the directed graph of
Fig. 9.10 starting at vertex e.
9.24. Show that when breadth-ﬁrst search is applied to an undirected graph
G, the edges of G will be classiﬁed as either tree edges or cross edges.
That is, there are no back edges or forward edges.
9.25. Show that when breadth-ﬁrst search is applied to a directed graph G,
the edges of G will be classiﬁed as tree edges, back edges or cross edges.
That is, unlike the case of depth-ﬁrst search, the search does not result
in forward edges.
9.26. Let G be a graph (directed or undirected), and let s be a vertex in G.
Modify Algorithm bfs so that it outputs the shortest path measured in
the number of edges from s to every other vertex.
9.27. Use depth-ﬁrst search to ﬁnd a spanning tree for the complete bipartite
graph K
3,3
. (See Sec. 3.3 for the deﬁnition of K
3,3
).
9.28. Use breadth-ﬁrst search to ﬁnd a spanning tree for the complete bipartite
graph K
3,3
. Compare this tree with the tree obtained in Exercise 9.27.
9.29. Suppose that Algorithm bfs is applied to an undirected graph G. Give
an algorithm that classiﬁes the edges of G as either tree edges or cross
edges.
9.30. Suppose that Algorithm bfs is applied to a directed graph G. Give an
algorithm that classiﬁes the edges of G as either tree edges, back edges
or cross edges.
9.31. Show that the time complexity of breadth-ﬁrst search when applied on
a graph with n vertices and m edges is Θ(n +m).
9.32. Design an eﬃcient algorithm to determine whether a given graph is bi-
partite (see Sec. 3.3 for the deﬁnition of a bipartite graph).
Bibliographic notes 273
9.33. Design an algorithm to ﬁnd a cycle of shortest length in a directed graph.
Here the length of a cycle is measured in terms of its number of edges.
9.34. Let G be a connected undirected graph, and T the spanning tree resulting
from applying breadth-ﬁrst search on G starting at vertex r. Prove or
disprove that the height of T is minimum among all spanning trees with
root r.
9.7 Bibliographic notes
Graph traversals are discussed in several books on algorithms, either sep-
arately or intermixed with other graph algorithms (see the bibliographic
notes of Chapter 1). Hopcroft and Tarjan (1973) were the ﬁrst to recognize
the algorithmic importance of depth-ﬁrst search. Several applications of
depth-ﬁrst search can be found in this paper and in Tarjan (1972). Al-
gorithm strongconnectcomp for the strongly connected components is
similar to the one by Sharir (1981). Tarjan (1972) contains an algorithm
for ﬁnding the strongly connected components that needs only one depth-
ﬁrst search traversal. Breadth-ﬁrst search was discovered independently by
Moore (1959) and Lee (1961).
274
PART 4
Complexity of Problems
275
276
277
In this part of the book, we turn our attention to the study of the com-
putational complexity of a problem as opposed to the cost of a particular
algorithm to solve that problem. We deﬁne the computational complex-
ity of a problem to be the computational complexity of the most eﬃcient
algorithm to solve that problem.
In Chapter 10, we study a class of problems known as NP-complete
problems. This class of problems encompasses numerous problems drawn
from many problem domains. These problems share the property that if
any one problem in the class is solvable in polynomial time, then all other
problems in the class are solvable in polynomial time. We have chosen to
cover this topic informally, in the sense that no speciﬁc model of compu-
tation is assumed. Instead, only the abstract notion of algorithm is used.
This makes it easy to the novice reader to comprehend the ideas behind NP-
completeness without missing with the details of a formal model (e.g. the
Turing machine). The most important point stressed in this chapter is to
study the standard technique of proving that a problem is NP-complete.
This is illustrated using several examples of NP-complete problems.
A more formal treatment of NP-completeness, as a special case of com-
pleteness in general, is postponed to Chapter 11. This chapter is somewhat
advanced and relies heavily on more than one variant of the Turing machine
model of computation. This chapter is concerned with the classiﬁcation of
problems based on the amount of time and space needed to solve a particu-
lar problem. First, the two variants of Turing machines, one for measuring
time and the other for space, are introduced. Next, the most prominent
time and space classes are deﬁned, and the relationships among them are
studied. This is followed by deﬁning the technique of transformation or
reduction in the context of Turing machines. The notion of completeness
in general is then addressed with the help of some examples. Finally, the
chapter closes with a preview of the polynomial time hierarchy.
Chapter 12 is concerned with establishing lower bounds for various prob-
lems. In this chapter, two models of computations are used for that pur-
pose. First, the decision tree model is used to establish lower bounds for
comparison-based problems like searching and sorting. Next, the more
powerful algebraic decision tree model is used to establish lower bounds
for some other problems. Some of these problems belong to the ﬁeld of
computational geometry, namely the convex hull problem, the closest pair
problem and the Euclidean minimum spanning tree problem.
278
Chapter 10
NP-complete Problems
10.1 Introduction
In the previous chapters, we have been working mostly with the design and
analysis of those algorithms for which the running time can be expressed
in terms of a polynomial of low degree, say 3. In this chapter, we turn our
attention to a class of problems for which no eﬃcient algorithms have been
found. Moreover, it is unlikely that an eﬃcient algorithm will someday
be discovered for any one of these problems. Let Π be any problem. We
say that there exists a polynomial time algorithm to solve problem Π if
there exists an algorithm for Π whose time complexity is O(n
k
), where n
is the input size and k is a nonnegative integer. It turns out that many of
the interesting real world problems do not fall into this category, as their
solution requires an amount of time that is measured in terms of exponential
and hyperexponential functions, e.g. 2
n
and n!. It has been agreed upon
in the computer science community to refer to those problems for which
there exist polynomial time algorithms as tractable, and those for which it
is unlikely that there exist polynomial time algorithms as intractable.
In this chapter, we will study a subclass of intractable problems, com-
monly referred to as the class of NP-complete problems. This class contains,
among many others, hundreds of well-known problems having the common
property that if one of them is solvable in polynomial time, then all the
others are solvable in polynomial time. Interestingly, many of these are
natural problems in the sense that they arise in real world applications.
Moreover, the running times of the existing algorithms to solve these prob-
lems are invariably measured in terms of hundreds or thousands of years
279
280 NP-complete Problems
for inputs of moderate size (see Table 1.1 on page 24).
When studying the theory of NP-completeness, it is easier to restate a
problem so that its solution has only two outcomes: yes or no. In this case,
the problem is called a decision problem. In contrast, an optimization prob-
lem is a problem that is concerned with the minimization or maximization
of a certain quantity. In the previous chapters, we have encountered numer-
ous optimization problems, like ﬁnding the minimum or maximum in a list
of elements, ﬁnding the shortest path in a directed graph and computing a
minimum cost spanning tree of an undirected graph. In the following, we
give three examples of how to formulate a problem as a decision problem
and an optimization problem.
Example 10.1 Let S be a sequence of real numbers. The ELEMENT
UNIQUENESS problem asks whether all the numbers in S are distinct.
Rephrased as a decision problem, we have
Decision problem: element uniqueness.
Input: A sequence S of integers.
Question: Are there two elements in S that are equal?
Stated as an optimization problem, we are interested in ﬁnding an element in
S of highest frequency. For instance, if S = 1, 5, 4, 5, 6, 5, 4, then 5 is of highest
frequency since it appears in the sequence 3 times, which is maximum. Let us call
this optimization version element count. This version can be stated as follows.
Optimization problem: element count.
Input: A sequence S of integers.
Output: An element in S of highest frequency.
This problem can be solved in optimal O(nlog n) time in the obvious way, which
means it is tractable.
Example 10.2 Given an undirected graph G = (V, E), a coloring of G using
k colors is an assignment of one of k colors to each vertex in V in such a way
that no two adjacent vertices have the same color. The coloring problem asks
whether it is possible to color an undirected graph using a speciﬁed number of
colors. Formulated as a decision problem, we have
Decision problem: coloring.
Input: An undirected graph G = (V, E) and a positive integer k ≥ 1.
Question: Is G k-colorable?, i.e., can G be colored using at most k colors?
This problem is intractable. If k is restricted to 3, the problem reduces to the
well-known 3-coloring problem, which is also intractable even when the graph
is planar.
An optimization version of this problem asks for the minimum number of colors
needed to color a graph in such a way that no two adjacent vertices have the
same color. This number, denoted by χ(G), is called the chromatic number of G.
Introduction 281
Optimization problem: chromatic number.
Input: An undirected graph G = (V, E).
Output: The chromatic number of G.
Example 10.3 Given an undirected graph G = (V, E), a clique of size k in
G, for some positive integer k, is a complete subgraph of G with k vertices. The
clique problem asks whether an undirected graph contains a clique of a speciﬁed
size. Rephrased as a decision problem, we have
Decision problem: clique.
Input: An undirected graph G = (V, E) and a positive integer k.
Question: Does G have a clique of size k?
The optimization version of this problem asks for the maximum number k such
that G contains a clique of size k, but no clique of size k + 1. We will call this
problem max-clique.
Optimization problem: max-clique.
Input: An undirected graph G = (V, E).
Output: A positive integer k, which is the maximum clique size in G.
If we have an eﬃcient algorithm that solves a decision problem, then
it can easily be modiﬁed to solve its corresponding optimization problem.
For instance, if we have an algorithm A that solves the decision problem for
graph coloring, we can ﬁnd the chromatic number of a graph G using binary
search and Algorithm A as a subroutine. Clearly, 1 ≤ χ(G) ≤ n, where n
is the number of vertices in G. Hence, the chromatic number of G can be
found using only O(log n) calls to algorithm A. Since we are dealing with
polynomial time algorithms, the log n factor is inconsequential. Because
of this reason, in the study of NP-complete problems, and computational
complexity or even computability in general, it is easier to restrict one’s
attention to decision problems.
It is customary in the study of NP-completeness, or computational com-
plexity in general, to adopt a formal model of computation such as the
Turing machine model of computation, as it makes the topic more formal
and the proofs more rigorous. In this chapter, however, we will work with
the abstract notion of “algorithm”, and will not attempt to formalize it by
associating it with any model of computation. A more formal treatment
that uses the Turing machine as a model of computation can be found in
Chapter 11.
282 NP-complete Problems
10.2 The Class P
Deﬁnition 10.1 Let A be an algorithm to solve a problem Π. We say
that A is deterministic if, when presented with an instance of the problem
Π, it has only one choice in each step throughout its execution. Thus, if A is
run again and again on the same input instance, its output never changes.
All algorithms we have covered in the previous chapters are determinis-
tic. The modiﬁer “deterministic” will mostly be dropped if it is understood
from the context.
Deﬁnition 10.2 The class of decision problems P consists of those deci-
sion problems whose yes/no solution can be obtained using a deterministic
algorithm that runs in polynomial number of steps, i.e., in O(n
k
) steps, for
some nonnegative integer k, where n is the input size.
We have encountered numerous such problems in the previous chap-
ters. Since in this chapter we are dealing with decision problems, we list
here some of the decision problems in the class P. The solutions to these
problems should be fairly easy.
sorting: Given a list of n integers, are they sorted in nondecreasing order?
set disjointness: Given two sets of integers, is their intersection empty?
shortest path: Given a directed graph G = (V, E) with positive weights
on its edges, two distinguished vertices s, t ∈ V and a positive integer k, is
there a path from s to t whose length is at most k?
2-coloring: Given an undirected graph G, is it 2-colorable?, i.e., can its
vertices be colored using only 2 colors such that no two adjacent vertices
are assigned the same color? Note that G is 2-colorable if and only if it is
bipartite, that is, if and only if it does not contain cycles of odd length (see
Sec. 3.3).
2-sat: Given a boolean expression f in conjunctive normal form (CNF),
where each clause consists of exactly two literals, is f satisﬁable? (See
Sec. 10.4.1).
We say that a class of problems ( is closed under complementation if
for any problem Π ∈ ( the complement of Π is also in (. For instance, the
complement of the 2-coloring problem can be stated as follows. Given
a graph G, is it not 2-colorable? Let us call this problem not-2-color.
We can show that it is in P as follows. Since 2-coloring is in P, there
The Class NP 283
is a deterministic algorithm A which when presented with a 2-colorable
graph halts and answers yes, and when presented with a graph that is not
2-colorable halts and answers no. We can simply design a deterministic
algorithm for the problem not-2-color by simply interchanging the yes
and no answers in Algorithm A. This, informally, proves the following
fundamental theorem:
Theorem 10.1 The class P is closed under complementation.
10.3 The Class NP
The class NP consists of those problems Π for which there exists a deter-
ministic algorithm A which, when presented with a claimed solution to an
instance of Π, will be able to verify its correctness in polynomial time. That
is, if the claimed solution leads to a yes answer, there is a way to verify
this solution in polynomial time.
In order to deﬁne this class less informally, we must ﬁrst deﬁne the
concept of a nondeterministic algorithm. On input x, a nondeterministic
algorithm consists of two phases:
(a) The guessing phase. In this phase, an arbitrary string of characters y
is generated. It may correspond to a solution to the input instance or not.
In fact, it may not even be in the proper format of the desired solution. It
may diﬀer from one run to another of the nondeterministic algorithm. It
is only required that this string be generated in a polynomial number of
steps, i.e., in O(n
i
) time, where n = [x[ and i is a nonnegative integer. In
many problems, this phase can be accomplished in linear time.
(b) The veriﬁcation phase. In this phase, a deterministic algorithm veriﬁes
two things. First, it checks whether the generated solution string y is in
the proper format. If it is not, then the algorithm halts with the answer
no. If, on the other hand, y is in the proper format, then the algorithm
continues to check whether it is a solution to the instance x of the problem.
If it is indeed a solution to the instance x, then it halts and answers yes;
otherwise it halts and answers no. It is also required that this phase be
completed in a polynomial number of steps, i.e., in O(n
j
) time, where j is
a nonnegative integer.
284 NP-complete Problems
Let A be a nondeterministic algorithm for a problem Π. We say that
A accepts an instance I of Π if and only if on input I there is a guess that
leads to a yes answer. In other words, A accepts I if and only if it is possible
on some run of the algorithm that its veriﬁcation phase will answer yes.
It should be emphasized that if the algorithm answers no, then this does
not mean that A does not accept its input, as the algorithm might have
guessed an incorrect solution.
As to the running time of a (nondeterministic) algorithm, it is simply
the sum of the two running times: the one for the guessing phase, and
that for the veriﬁcation phase. So it is O(n
i
) + O(n
j
) = O(n
k
), for some
nonnegative integer k.
Deﬁnition 10.3 The class of decision problems NP consists of those
decision problems for which there exists a nondeterministic algorithm that
runs in polynomial time.
Example 10.4 Consider the problem coloring. We show that this problem
belongs to the class NP in two ways.
(1) The ﬁrst method is as follows. Let I be an instance of the problem coloring.
Let s be a claimed solution to I. It is easy to construct a deterministic algorithm
that tests whether s is indeed a solution to I. It follows by our informal deﬁnition
of the class NP that the problem coloring belongs to the class NP.
(2) The second method is to construct a nondeterministic algorithm for this prob-
lem. An algorithm A can easily be constructed that does the following when
presented with an encoding of a graph G. First, A “guesses” a solution by gener-
ating an arbitrary assignment of the colors to the set of vertices. Next, A veriﬁes
that the guess is a valid assignment. If it is a valid assignment, then A halts and
answers yes; otherwise it halts and answers no. First, note that according to the
deﬁnition of a nondeterministic algorithm, A answers yes only if the answer to
the instance of the problem is yes. Second, regarding the operation time needed,
A spends no more than polynomial time in both the guessing and veriﬁcation
phases.
To this end, we have the following distinction between the two important
classes P and NP:
• P is the class of decision problems that we can decide or solve using
a deterministic algorithm that runs in polynomial time.
• NP is the class of decision problems that we can check or verify their
solution using a deterministic algorithm that runs in polynomial
time.
NP-complete Problems 285
10.4 NP-complete Problems
The term “NP-complete” denotes the subclass of decision problems in NP
that are hardest in the sense that if one of them is proven to be solvable
by a polynomial time deterministic algorithm, then all problems in NP are
solvable by a polynomial time deterministic algorithm, i.e., NP = P. For
proving that a problem is NP-complete, we need the following deﬁnition.
Deﬁnition 10.4 Let Π and Π

be two decision problems. We say that Π
reduces to Π

in polynomial time, symbolized as Π ∝
poly
Π

, if there exists
a deterministic algorithm A that behaves as follows. When A is presented
with an instance I of problem Π, it transforms it into an instance I

of
problem Π

such that the answer to I is yes if and only if the answer to
I

is yes. Moreover, this transformation must be achieved in polynomial
time.
Deﬁnition 10.5 A decision problem Π is said to be NP-hard if for every
problem Π

in NP, Π

∝
poly
Π.
Deﬁnition 10.6 A decision problem Π is said to be NP-complete if
(1) Π is in NP, and
(2) for every problem Π

may not be
in NP.
10.4.1 The satisﬁability problem
Given a boolean formula f, we say that it is in conjunctive normal form
(CNF) if it is the conjunction of clauses. A clause is the disjunction of
literals, where a literal is a boolean variable or its negation. An example of
such a formula is
f = (x
1
∨ x
2
) ∧ (x
1
∨ x
3
∨ x
4
∨ x
5
) ∧ (x
1
∨ x
3
∨ x
4
).
A formula is said to be satisﬁable if there is a truth assignment to its
variables that makes it true. For example, the above formula is satisﬁable,
since it evaluates to true under any assignment in which both x
1
and x
3
are set to true.
286 NP-complete Problems
Decision problem: satisfiability.
Input: A CNF boolean formula f.
Question: Is f satisﬁable?
The satisﬁability problem was the ﬁrst problem proven to be NP-
complete. Being the ﬁrst NP-complete problem, there was no other NP-
complete problem that reduces to it. Therefore, the proof was to show that
all problems in the class NP can be reduced to it in polynomial time. In
other words, the essence of the proof is to show that any problem in NP can
be solved by a polynomial time algorithm that uses the satisﬁability prob-
lem as a subroutine that is invoked by the algorithm exactly once. The proof
consists of constructing a boolean formula f in conjunctive normal form for
an instance I of Π such that there is a truth assignment that satisﬁes f if
and only if a nondeterministic algorithm A for the problem Π accepts the
instance I. f is constructed so that it “simulates” the computation of A on
instance I. This, informally, implies the following fundamental theorem:
Theorem 10.2 satisfiability is NP-complete.
The following theorem states that the reducibility relation ∝
P
is tran-
sitive. This is necessary to show that other problems are NP-complete as
well. We explain this as follows. Suppose that for some problem Π in NP
we can prove that satisfiability reduces to Π in polynomial time. By the
above theorem, all problems in NP reduce to satisfiability in polynomial
time. Consequently, if the reducibility relation ∝
P
is transitive, then this
implies that all problems in NP reduce to Π in polynomial time.
Theorem 10.3 Let Π, Π

and Π

be three decision problems such that
Π ∝
poly
Π

and Π

∝
poly
Π

. Then Π ∝
poly
Π

.
Proof. Let A be an algorithm that realizes the reduction Π ∝
poly
Π

in
p(n) steps for some polynomial p. Let B be an algorithm that realizes the
reduction Π

∝
poly
Π

in q(n) steps for some polynomial q. Let x be an
input to A of size n. Clearly, the size of the output of algorithm A when
presented with input x cannot exceed cp(n), as the algorithm can output
at most c symbols in each step of its execution for some positive integer
c > 0. If algorithm B is presented with an input of size p(n) or less, its
running time is, by deﬁnition, O(q(cp(n))) = O(r(n)) for some polynomial
r. It follows that the reduction from Π to Π

followed by the reduction from
Π

to Π

is a polynomial time reduction from Π to Π

. ⁄
NP-complete Problems 287
Corollary 10.1 If Π and Π

are two problems in NP such that
Π

∝
poly
Π, and Π

is NP-complete, then Π is NP-complete.
Proof. Since Π

is NP-complete, every problem in NP reduces to Π

in
polynomial time. Since Π

∝
poly
Π, then by Theorem 10.3, every problem
in NP reduces to Π in polynomial time. It follows that Π is NP-complete.
⁄
By the above corollary, to prove that a problem Π is NP-complete, we
only need to show that
(1) Π ∈ NP, and
(2) there is an NP-complete problem Π

such that Π

∝
poly
Π.
Example 10.5 Consider the following two problems.
(1) The problem Hamiltonian cycle: Given an undirected graph G = (V, E),
does it have a Hamiltonian cycle, i.e., a cycle that visits each vertex exactly once?
(2) The problem traveling salesman: Given a set of n cities with their intercity
distances, and an integer k, does there exist a tour of length at most k? Here, a
tour is a cycle that visits each city exactly once.
It is well known that the problem Hamiltonian cycle is NP-complete. We will
use this fact to show that the problem traveling salesman is also NP-complete.
The ﬁrst step in the proof is to show that traveling salesman is in NP.
This is very simple, since a nondeterministic algorithm can start by guessing a
sequence of cities, and then veriﬁes that this sequence is a tour. If this is the case,
it then continues to see if the length of the tour is at most k, the given bound.
The second step is to show that Hamiltonian cycle can be reduced to
traveling salesman in polynomial time, i.e.,
Hamiltonian cycle ∝
poly
traveling salesman.
Let G be any arbitrary instance of Hamiltonian cycle. We construct a weighted
graph G

and a bound k such that G has a Hamiltonian cycle if and only if G

has a tour of total length at most k. Let G = (V, E). We let G

= (V, E

) be the
complete graph on the set of vertices V , i.e.,
E

= ¦(u, v) [ u, v ∈ V ¦.
Next, we assign a length to each edge in E

as follows.
l(e) =

1 if e ∈ E
n if e / ∈ E,
288 NP-complete Problems
where n = [V [. Finally, we assign k = n. It is easy to see from the construction
that G has a Hamiltonian cycle if and only if G

has a tour of length exactly n.
It should be emphasized that the assignment k = n is part of the reduction.
10.4.2 vertex cover, independent set and clique problems
In this section, we prove the NP-completeness of three famous problems in
graph theory.
clique: Given an undirected graph G = (V, E) and a positive integer k,
does G contain a clique of size k? (recall that a clique in G of size k is a
complete subgraph of G on k vertices).
vertex cover: Given an undirected graph G = (V, E) and a positive
integer k, is there a subset C ⊆ V of size k such that each edge in E is
incident to at least one vertex in C?
independent set: Given an undirected graph G = (V, E) and a positive
integer k, does there exist a subset S ⊆ V of k vertices such that for each
pair of vertices u, w ∈ S, (u, w) / ∈ E?
It is easy to show that all these three problems are indeed in NP. In what
follows we give reductions that establish their NP-completeness.
satisfiability ∝
poly
clique
Given an instance of satisfiability f = C
1
∧C
2
∧. . . ∧C
m
with m clauses
and n boolean variables x
1
, x
2
, . . . , x
n
, we construct a graph G = (V, E),
where V is the set of all occurrences of the 2n literals (recall that a literal
is a boolean variable or its negation), and
E = ¦(x
i
, x
j
) [ x
i
and x
j
are in two diﬀerent clauses and x
i
= x
j
¦.
It is easy to see that the above construction can be accomplished in poly-
nomial time.
Example 10.6 An example of the reduction is provided in Fig. 10.1. Here
the instance of satisfiability is
f = (x ∨ y ∨ z) ∧ (x ∨ y) ∧ (x ∨ y ∨ z).
Lemma 10.1 f is satisﬁable if and only if G has a clique of size m.
NP-complete Problems 289
j
a .
a
j j
a .
Fig. 10.1 Reducing satisfiability to clique.
Proof. A clique of size m corresponds to an assignment of true to m
literals in m diﬀerent clauses. An edge between two literals a and b means
that there is no contradiction when both a and b are assigned the value
true. It follows that f is satisﬁable if and only if there is a noncontradictory
assignment of true to m literals in m diﬀerent clauses if and only if G has
a clique of size m. ⁄
satisfiability ∝
poly
vertex cover
Given an instance I of satisfiability, we transform it into an instance I

as follows.
(1) For each boolean variable x
i
in f, G contains a pair of vertices x
i
and x
i
joined by an edge.
(2) For each clause C
j
containing n
j
literals, G contains a clique C
j
of size
n
j
.
(3) For each vertex w in C
j
, there is an edge connecting w to its corre-
sponding literal in the vertex pairs (x
i
, x
i
) constructed in part (1). Call
these edges connection edges.
(4) Let k = n +
¸
m
j=1
(n
j
−1).
It is easy to see that the above construction can be accomplished in
polynomial time.
290 NP-complete Problems
Example 10.7 An example of the reduction is provided in Fig. 10.2. Here
the instance I is the formula
f = (x ∨ y ∨ z) ∧ (x ∨ y).
. . j a
j a
Fig. 10.2 Reducing satisfiability to vertex cover.
It should be emphasized that the instance I

is not only the ﬁgure shown;
it also includes the integer k = 3 + 2 + 1 = 6. A boolean assignment of x =
true, y = true and z = false satisﬁes f. This assignment corresponds to the 6
covering vertices shown shaded in the ﬁgure.
Lemma 10.2 f is satisﬁable if and only if the constructed graph has a
vertex cover of size k.
Proof. ⇒: If x
i
is assigned true, add vertex x
i
to the vertex cover; oth-
erwise add x
i
to the vertex cover. Since f is satisﬁable, in each clique
C
j
there is a vertex u whose corresponding literal v has been assigned the
value true, and thus the connection edge (u, v) is covered. Therefore, add
the other n
j
−1 vertices in each clique C
j
to the vertex cover. Clearly, the
size of the vertex cover is k = n +
¸
m
j=1
(n
j
−1).
⇐: Suppose that the graph can be covered with k vertices. At least one
vertex of each edge (x
i
, x
i
) must be in the cover. We are left with k −n =
¸
m
j=1
(n
j
− 1) vertices. It is not hard to see that any cover of a clique of
size n
j
must have exactly n
j
− 1 vertices. So, in each clique, the cover
must include all its vertices except the one that is incident to a connection
edge that is covered by a vertex in some vertex pair (x
i
, x
i
). To see that
f is satisﬁable, for each vertex x
i
, if it is in the cover then let x
i
= true;
otherwise (if x
i
is in the cover), let x
i
= false. Thus, in each clique, there
NP-complete Problems 291
must be one vertex which is connected to a vertex x
i
or x
i
, which is assigned
the value true since it is in the cover. It follows that each clause has at
least one literal whose value is true, i.e., f is satisﬁable. ⁄
vertex cover ∝
poly
independent set
The transformation from vertex cover to independent set is straight-
forward. The following lemma provides the reduction:
Lemma 10.3 Let G = (V, E) be a connected undirected graph. Then,
S ⊆ V is an independent set if and only if V −S is a vertex cover in G.
Proof. Let e = (u, v) be any edge in G. S is an independent set if and
only if at least one of u or v is in V −S, i.e., V −S is a vertex cover in G.
⁄
A simple reduction from vertex cover to clique is left as an exer-
cise (Exercise 10.16). A reduction from independent set to clique is
straightforward since a clique in a graph G is an independent set in G, the
complement of G. Thus, we have the following theorem:
Theorem 10.4 The problems vertex cover, independent set and
clique are NP-complete.
Proof. It is fairly easy to show that these problems are in the class NP.
The above reductions whose proofs are given in Lemmas 10.1, 10.2 and 10.3
complete the proof. ⁄
10.4.3 More NP-complete Problems
The following is a list of additional NP-complete problems.
(1) 3-sat. Given a boolean formula f in conjunctive normal form such that
each clause consists of exactly three literals, is f satisﬁable?
(2) 3-coloring. Given an undirected graph G = (V, E), can G be colored
using three colors? This problem is a special case of the more general
problem coloring stated on page 280, which is known to be NP-complete.
(3) 3-dimensional matching. Let X, Y and Z be pairwise disjoint sets of
size k each. Let W be the set of triples ¦(x, y, z) [ x ∈ X, y ∈ Y, z ∈ Z¦.
Does there exist a perfect matching M of W? That is, does there exist
292 NP-complete Problems
a subset M ⊆ W of size k such that no two triplets in M agree in any
coordinate? The corresponding two-dimensional matching problem is the
regular perfect bipartite matching problem (see Chapter 17).
(4) Hamiltonian path. Given an undirected graph G = (V, E), does it
contain a simple open path that visits each vertex exactly once?
(5) partition. Given a set S of n integers, is it possible to partition S
into two subsets S
1
and S
2
so that the sum of the integers in S
1
is equal to
the sum of the integers in S
2
?
(6) knapsack. Given n items with sizes s
1
, s
2
, . . . , s
n
and values
v
1
, v
2
, . . . , v
n
, a knapsack capacity C and a constant integer k, is it possible
to ﬁll the knapsack with some of these items whose total size is at most C
and whose total value is at least k? This problem can be solved in time
Θ(nC) using dynamic programming (Theorem 7.3).
(7) bin packing. Given n items with sizes s
1
, s
2
, . . . , s
n
, a bin capacity C
and a positive integer k, is it possible to pack the n items using at most k
bins?
(8) set cover. Given a set X, a family T of subsets of X and an integer
k between 1 and [T[, do there exist k subsets in T whose union is X?
(9) multiprocessor scheduling. Given n jobs J
1
, J
2
, . . . , J
n
each hav-
ing a run time t
i
, a positive integer m (number of processors), and a ﬁnish-
ing time T, can these jobs be scheduled on m identical processors so that
their ﬁnishing time is at most T? The ﬁnishing time is deﬁned to be the
maximum execution time among all the m processors.
(10) longest path. Given a weighted graph G = (V, E), two distinguished
vertices s, t ∈ V and a positive integer c, is there a simple path in G from
s to t of length c or more?
10.5 The Class co-NP
The class co-NP consists of those problems whose complements are in NP.
One might suspect that the class co-NP is comparable in hardness to the
class NP. It turns out, however, that this is highly unlikely, which supports
the conjecture that co-NP = NP. Consider, for example, the complement
The Class co-NP 293
of traveling salesman: Given n cities with their intercity distances, is
it the case that there does not exist any tour of length k or less? It seems
that there is no nondeterministic algorithm that solves this problem with-
out exhausting all the (n − 1)! possibilities. As another example, consider
the complement of satisfiability: Given a formula f, is it the case that
there is no assignment of truth values to its boolean variables that satisﬁes
f? In other words, is f unsatisﬁable? There does not seem to be a non-
deterministic algorithm that solves this problem without inspecting all the
2
n
assignments, where n is the number of boolean variables in f.
Deﬁnition 10.7 A problem Π is complete for the class co-NP if
(1) Π is in co-NP, and
(2) for every problem Π

in co-NP, Π

∝
poly
Π.
Let some deterministic algorithm A realize a reduction from one prob-
lem Π

to another problem Π, both in NP. Recall that, by deﬁnition of
reduction, A is deterministic and runs in polynomial time. Therefore, by
Theorem 10.1, A is also a reduction from Π

to Π, where Π and Π

are the
complements of Π and Π

, respectively. This implies the following theorem:
Theorem 10.5 A problem Π is NP-complete if and only if its comple-
ment, Π, is complete for the class co-NP.
In particular, since satisfiability is NP-complete, the complement of
satisfiability is complete for the class co-NP. It is not known whether the
class co-NP is closed under complementation. It follows, however, that the
complement of satisfiability is in NP if and only if NP is closed under
complementation.
A CNF formula f is unsatisﬁable if and only if its negation is a tautology
(A formula f is called a tautology if f is true under all truth assignments to
its boolean variables). The negation of a CNF formula C
1
∧ C
2
∧ . . . ∧ C
k
,
where C
i
= (x
1
∨ x
2
∨ . . . ∨ x
m
i
), for all i, 1 ≤ i ≤ k, can be converted
into a disjunctive normal form (DNF) formula C

1
∨ C

2
∨ . . . ∨ C

k
, where
C

i
= (y
1
∧ y
2
∧ . . . ∧ y
m
i
), for all i, 1 ≤ i ≤ k, using the identities
(C
1
∧ C
2
∧ . . . ∧ C
k
) = (C
1
∨ C
2
∨ . . . ∨ C
k
)
and
(x
1
∨ x
2
∨ . . . ∨ x
m
i
) = (x
1
∧ x
2
∧ . . . ∧ x
m
i
).
294 NP-complete Problems
The resulting DNF formula is a tautology if and only if the negation of the
CNF formula is a tautology. Therefore, we have the following theorem:
Theorem 10.6 The problem tautology: Given a formula f in DNF,
is it a tautology? is complete for the class co-NP.
It follows that
• tautology is in P if and only if co-NP = P, and
• tautology is in NP if and only if co-NP = NP.
10.6 The Class NPI
The following theorem is fundamental. Its simple proof is left as an exercise
(Exercise 10.30).
Theorem 10.7 If a problem Π and its complement Π are NP-complete,
then co-NP = NP.
In other words, If both a problem Π and its complement are NP-
complete, then the class NP is closed under complementation. As dis-
cussed before, this is highly unlikely, and it is an open question. In fact, it
is stronger than the NP = P question. The reason is that if we can prove
that co-NP = NP, then it follows immediately that NP = P. For suppose
it has been proven that co-NP = NP, and assume for the sake of contra-
diction that NP = P. Then, substituting P for NP in the proven result,
we obtain co-P = P. But this contradicts the fact that P is closed under
complementation (Theorem 10.1). This contradiction implies that NP = P.
There are some problems in NP whose complements are also in NP.
Two such problems are the following:
• prime number: Given an integer k ≥ 2, is k a prime number?
• composite number: Given an integer k ≥ 4, are there two inte-
gers p, q ≥ 2 such that k = pq?
To show that composite number is in NP is trivial. However, showing
that prime number is in NP is not straightforward, and we will not go
into its details here. Now, can problem prime number be NP-complete?
This is highly unlikely. For if prime number is NP-complete, then by
The Relationships Between the Four Classes 295
Theorem 10.5, its complement composite number will be complete for
the class co-NP. This means any problem in co-NP reduces to composite
number. But composite number is in NP. This implies that all problems
in co-NP are in NP or, in other words, co-NP = NP. And, as stated
above, this is highly unlikely. Hence, it is unlikely that problems like prime
number or composite number turn out to be NP-complete.
Due to the very slim probability that these two problems are NP-
complete, and also due to the belief that they are not in P, another class
of problems has been added to accommodate such problems. This class is
called NPI, which stands for “NP-Intermediate”. The linear program-
ming problem was thought to belong to this class until it was shown to be
solvable in polynomial time.
10.7 The Relationships Between the Four Classes
Figure 10.3 shows the relationships between the four classes we have dis-
cussed in this chapter. From the ﬁgure, it is clear that the class NPI is
closed under complementation and that P ⊆ NPI.
complete
co-NP-
NP-
complete
TAUTOLOGY
UNSAT
SORTING
LINEAR
PROGRAMMING
TSP SAT
CLIQUE
VC IS
COMPOSITE
PRIME
MATCHING
CO-NP
NP
NPI
P
Fig. 10.3 The relationships between the four complexity classes.
296 NP-complete Problems
10.8 Exercises
10.1. Give an eﬃcient algorithm to solve the decision version of the sorting
stated on page 282. What is the time complexity of your algorithm?
10.2. Give an eﬃcient algorithm to solve the problem set disjointness stated
on page 282. What is the time complexity of your algorithm?
10.3. Design a polynomial time algorithm for the problem 2-coloring deﬁned
on page 282. (Hint: Color the ﬁrst vertex white, all adjacent vertices
black, etc).
10.4. Design a polynomial time algorithm for the problem 2-sat deﬁned on
page 282.
10.5. Let I be an instance of the problem coloring, and let s be a claimed
solution to I. Describe a deterministic algorithm to test whether s is a
solution to I.
10.6. Design a nondeterministic algorithm to solve the problem satisfiabil-
ity.
10.7. Design a nondeterministic algorithm to solve the problem traveling
salesman.
10.8. Show that P ⊆ NP.
10.9. Let Π
1
and Π
2
be two problems such that Π
1
∝
poly
Π
2
. Suppose that
problem Π
2
can be solved in O(n
k
) time and the reduction can be done
in O(n
j
) time. Show that problem Π
1
can be solved in O(n
jk
) time.
10.10. Given that the Hamiltonian cycle problem for undirected graphs is NP-
complete, show that the Hamiltonian cycle problem for directed graphs
is also NP-complete.
10.11. Show that the problem bin packing is NP-complete, assuming that the
problem partition is NP-complete.
10.12. Let Π
1
and Π
2
be two NP-complete problems. Prove or disprove that
Π
1
∝
poly
Π
2
.
10.13. Give a polynomial time algorithm to ﬁnd a clique of size k in a given
undirected graph G = (V, E) with n vertices. Here k is a ﬁxed positive
integer. Does this contradict the fact that the problem clique is NP-
complete? Explain.
10.14. Consider the following instance of satisfiability:
(x
1
∨ x
2
∨ x
3
) ∧ (x
1
∨ x
3
) ∧ (x
2
∨ x
3
) ∧ (x
1
∨ x
2
)
(a) Following the reduction method from satisfiability to clique,
transform the above formula into an instance of clique for which
the answer is yes if and only if the the above formula is satisﬁable.
Exercises 297
(b) Find a clique of size 4 in your graph and convert it into a satisfying
assignment for the formula given above.
10.15. Consider the formula f given in Exercise 10.14.
(a) Following the reduction method from satisfiability to vertex
cover, transform f into an instance of vertex cover for which
the answer is yes if and only if f is satisﬁable.
(b) Find a vertex cover in your graph and convert it into a satisfying
assignment for f.
10.16. The NP-completeness of the problem clique was shown by reducing
satisfiability to it. Give a simpler reduction from vertex cover to
clique.
10.17. Show that any cover of a clique of size n must have exactly n−1 vertices.
10.18. Show that if one can devise a polynomial time algorithm for the problem
satisfiability then NP = P (see Exercise 10.9).
10.19. In Chapter 7 it was shown that the problem knapsack can be solved in
time Θ(nC), where n is the number of items and C is the knapsack ca-
pacity. However, it was mentioned in this chapter that it is NP-complete.
Is there any contradiction? Explain.
10.20. When showing that an optimization problem is not harder than its deci-
sion problem version, it was justiﬁed by using binary search and an algo-
rithm for the decision problem in order to solve the optimization version.
Will the justiﬁcation still be valid if linear search is used instead of binary
search? Explain. (Hint: Consider the problem traveling salesman).
10.21. Prove that if an NP-complete problem Π is shown to be solvable in
polynomial time, then NP = P (see Exercises 10.9 and 10.18).
10.22. Prove that NP = P if and only if for some NP-complete problem Π,
Π ∈ P.
10.23. Is the problem longest path NP-complete when the path is not re-
stricted to be simple? Prove your answer.
10.24. Is the problem longest path NP-complete when restricted to directed
acyclic graphs? Prove your answer. (See Exercises 10.23 and 7.34).
10.25. Show that the problem of ﬁnding a shortest simple path between two
vertices s and t in a directed or undirected graph is NP-complete if the
weights are allowed to be negative.
10.26. Show that the problem set cover is NP-complete by reducing the prob-
lem vertex cover to it.
10.27. Show that the problem 3-sat is NP-complete.
10.28. Show that the problem 3-coloring is NP-complete.
298 NP-complete Problems
10.29. Compare the diﬃculty of the problem tautology to satisfiability.
What does this imply about the diﬃculty of the class co-NP.
10.30. Prove Theorem 10.7.
10.9 Bibliographic notes
The study of NP-completeness started with two papers. The ﬁrst was
the seminal paper of Cook (1971) in which the problem satisfiability
was the ﬁrst problem shown to be NP-complete. The second was Karp
(1972) in which a list of 24 problems were shown to be NP-complete. Both
Stephen Cook and Richard Karp have won the ACM Turing awards and
their Turing award lectures were published in Cook (1983) and Karp (1986).
Garey and Johnson (1979) provides comprehensive coverage of the theory
of NP-completeness and covers the four basic complexity classes introduced
in this chapter. Their book contains the proof that satisfiability is NP-
complete and a list of several hundred NP-complete problems. One of the
most famous of the open problems to be resolved is linear programming.
This problem has been proven to be solvable in polynomial time using
the ellipsoid method (Khachiyan (1979)). It has received much attention,
although its practical signiﬁcance is yet to be determined. An introduction
to the theory of NP-completeness can also be found in Aho, Hopcroft and
Ullman (1974) and Hopcroft and Ullman (1979).
Chapter 11
Introduction to Computational
Complexity
11.1 Introduction
Computational complexity is concerned with the classiﬁcation of problems
based on the amount of time, space or any other resource needed to solve
a problem such as the number of processors and communication cost. In
this chapter, we review some of the basic concepts in this ﬁeld, and conﬁne
our attention to the two classical resource measures: time and space.
11.2 Model of Computation: the Turing Machine
When studying computational complexity, a universal computing device is
required for the classiﬁcation of computational problems. It turns out that
most, if not all, of the results are robust and are invariant under diﬀer-
ent models of computations. In this chapter, we will choose the Turing
machine as our model of computation. In order to measure the amount
of time and space needed to solve a problem, it will be much easier to
consider those problems whose solution output is either yes or no. A
problem of this type is called a decision problem (see Sec. 10.1). The
set of instances of a decision problem is partitioned into two sets: those
instances for which the answer is yes and those for which the answer is
no. We can encode such problems as languages. An alphabet Σ is a ﬁnite
set of symbols. A language L is simply a subset of the set of all ﬁnite
length strings of symbols chosen from Σ, denoted by Σ
∗
. For example, a
graph G = (V, E), where V = ¦1, 2, . . . , n¦ can be encoded by the string
w(G) = (x
11
, x
12
, . . . , x
1n
)(x
21
, x
22
, . . . , x
2n
) . . . (x
n1
, x
n2
, . . . , x
nn
), where
299
300 Introduction to Computational Complexity
x
ij
= 1 if (i, j) ∈ E and 0 otherwise. Thus the encoded graph w(G) is a
string of symbols over the ﬁnite alphabet ¦0, 1, (, )¦.
The standard Turing machine has only one worktape, which is divided
into separate cells. Each cell of the worktape contains one symbol from some
ﬁnite alphabet Σ. The Turing machine has the ability to read and rewrite
the symbol contained in the cell of the worktape currently scanned by its
worktape head. The worktape head moves either one cell to the left, one
cell to the right, or it remains on the current cell at each step. The actions
of the Turing machine are speciﬁed by its ﬁnite state control . A Turing
machine at any moment of time is in some state. For the current state and
for the current scanned symbol on the worktape, the ﬁnite state control
speciﬁes which actions are possible: It speciﬁes which one of the states to
enter next, which symbol to print on the scanned cell of the worktape, and
in what way to move the worktape head.
11.3 k-tape Turing machines and time complexity
Since the standard Turing machine as described in the previous section can
move its worktape head one cell per step, it clearly needs n steps to move
the head n cells. In order to make an appropriate model of Turing machine
adequately measure the amount of time used by an algorithm, we need to
allow for more than one worktape. A k-tape Turing machine, for some
k ≥ 1, is just the natural extension of the one-tape Turing machine. It
has k worktapes instead of just one, and it has k worktape heads. Each
worktape head is associated with one of the worktapes and the heads can
move independently from one another.
Deﬁnition 11.1 A (nondeterministic) k-tape Turing machine is a 6-tuple
M = (S, Σ, Γ, δ, p
0
, p
f
), where
(1) S is a ﬁnite set of states,
(2) Γ is a ﬁnite set of tape symbols which includes the special symbol
B (the blank symbol),
(3) Σ ⊆ Γ −¦B¦, the set of input symbols,
(4) δ, the transition function, is a function that maps elements of SΓ
k
into ﬁnite subsets of S ((Γ −¦B¦) ¦L, P, R¦)
k
,
(5) p
0
∈ S, the initial state, and
(6) p
f
∈ S, the ﬁnal or accepting state.
k-tape Turing machines and time complexity 301
Note that we have assumed without loss of generality that there is only
one ﬁnal state. A k-tape Turing machine M = (S, Σ, Γ, δ, p
0
, p
f
) is de-
terministic if for every p ∈ S and for every a
1
, a
2
, . . . , a
k
∈ Γ, the set
δ(p, a
1
, a
2
, . . . , a
k
) contains at most one element.
Deﬁnition 11.2 Let M = (S, Σ, Γ, δ, p
0
, p
f
) be a k-tape Turing machine.
A conﬁguration of M is a (k + 1)-tuple
K = (p, w
11
↑w
12
, w
21
↑w
22
, . . . , w
k1
↑w
k2
),
where p ∈ S and w
j1
↑w
j2
is the contents of the j-th tape of M, 1 ≤ j ≤ k.
Here, the head of the jth tape is pointing to the ﬁrst symbol in the
string w
j2
. If w
j1
is empty, then the head is pointing to the ﬁrst nonblank
symbol on the tape. If w
j2
is empty, then the head is pointing to the
ﬁrst blank symbol after the string w
j1
. Both w
j1
and w
j2
may be empty,
indicating that the tape is empty. This is the case in the beginning of the
computation, where all tapes except possibly the input tape are empty.
Thus, the initial conﬁguration is denoted by
(p
0
, ↑x, ↑B, . . . , ↑B),
where x is the initial input. The set of ﬁnal or accepting conﬁgurations is
the set of all conﬁgurations
(p
f
, w
11
↑w
12
, w
21
↑w
22
, . . . , w
k1
↑w
k2
).
Deﬁnition 11.3 A computation by a Turing machine M on input x is
a sequence of conﬁgurations K
1
, K
2
, . . . , K
t
, for some t ≥ 1, where K
1
is
the initial conﬁguration, and for all i, 2 ≤ i ≤ t, K
i
results from K
i−1
in
one move of M. Here, t is referred to as the length of the computation.
If K
t
is a ﬁnal conﬁguration, then the computation is called an accepting
computation.
Deﬁnition 11.4 The time taken by a Turing machine M on input x,
denoted by T
M
(x), is deﬁned by:
(1) If there is an accepting computation of M on input x, then T
M
(x)
is the length of the shortest accepting computation, and
(2) If there is no accepting computation of M on input x, then
T
M
(x) = ∞.
302 Introduction to Computational Complexity
Let L be a language and f a function from the set of nonnegative in-
tegers to the set of nonnegative integers. We say that L is in DTIME(f)
(resp. NTIME(f)) if there exists a deterministic (resp. nondeterminis-
tic) Turing machine M that behaves as follows. On input x, if x ∈ L
then T
M
(x) ≤ f([x[); otherwise T
M
(x) = ∞. Similarly, we may deﬁne
DTIME(n
k
), NTIME(n
k
) for any k ≥ 1. The two classes P and NP dis-
cussed in Chapter 10 can now be deﬁned formally as follows.
P = DTIME(n) ∪ DTIME(n
2
) ∪ DTIME(n
3
) ∪ . . . ∪ DTIME(n
k
) ∪ . . .
and
NP = NTIME(n) ∪ NTIME(n
2
) ∪ NTIME(n
3
) ∪ . . . ∪ NTIME(n
k
) ∪ . . .
In other words, P is the set of all languages recognizable in polynomial time
using a deterministic Turing machine, and NP is the set of all languages
recognizable in polynomial time using a nondeterministic Turing machine.
We have seen many examples of problems in the class P in earlier chapters.
We have also encountered several problems that belong to the class NP in
Chapter 10. There are also other important time complexity classes, two
of them are
DEXT =
¸
c≥0
DTIME(2
cn
), NEXT =
¸
c≥0
NTIME(2
cn
),
EXPTIME =
¸
c≥0
DTIME(2
n
c
) NEXPTIME =
¸
c≥0
NTIME(2
n
c
).
Example 11.1 Consider the following 1-tape Turing machine M that recog-
nizes the language L = ¦a
n
b
n
[ n ≥ 1¦. Initially its tape contains the string a
n
b
n
.
M repeats the following step until all symbols on the tape have been marked, or
M cannot move its tape head. M marks the leftmost unmarked symbol if it is an
a and then moves its head all the way to the right and marks the rightmost un-
marked symbol if it is a b. If the number of a’s is equal to the number of b’s, then
all symbols on the tape will eventually be marked, and hence M will enter the
accepting state. Otherwise, either the number of a’s is less than or greater than
the number of b’s. If the number of a’s is less than the number of b’s, then after
all the a’s have been marked, and after marking the last b, the leftmost symbol
is a b, and hence M will not be able to move its tape head. This will also be the
Oﬀ-line Turing machines and space complexity 303
case if the number of a’s is greater than the number of b’s. It is easy to see that
if the input string is accepted, then the number of moves of the tape head is less
than or equal to cn
2
for some constant c > 0. It follows that L is in DTIME(n
2
).
11.4 Oﬀ-line Turing machines and space complexity
For an appropriate measure of space, we need to separate the space used
to store computed information. For example, to say that a Turing machine
uses only log n| of its worktape cells is possible only if we separate the
input string and we do not count the n cells used to store the input string
of length n. For this reason, our model of a Turing machine that will be
used to measure space complexity will have a separate read-only input tape.
The Turing machine is not permitted to rewrite the symbols that it scans
on the input tape. This version of Turing machines is commonly referred
to as an oﬀ-line Turing machine. The diﬀerence between a k-tape Turing
machine and an oﬀ-line Turing machine is that an oﬀ-line Turing machine
has exactly two tapes: a read-only input tape and a read-write worktape.
Deﬁnition 11.5 A (nondeterministic) oﬀ-line Turing machine is a 6-
tuple M = (S, Σ, Γ, δ, p
0
, p
f
), where
(1) S is a ﬁnite set of states,
(2) Γ is a ﬁnite set of tape symbols, which includes the special symbol
B (the blank symbol),
(3) Σ ⊆ Γ − ¦B¦ is the set of input symbols; it contains two special
symbols # and $ (the left endmarker and the right endmarker,
respectively).
(4) δ, the transition function, is a function that maps elements of S
ΣΓ into ﬁnite subsets of S ¦L, P, R¦ (Γ −¦B¦) ¦L, P, R¦,
(5) p
0
∈ S is the initial state, and
(6) p
f
∈ S is the ﬁnal or accepting state.
Note that we have assumed without loss of generality that there is only
one ﬁnal state. The input is presented to an oﬀ-line Turing machine in
its read-only tape enclosed by the endmarkers, $ and #, and it is never
changed. In the case of oﬀ-line Turing machines, a conﬁguration is deﬁned
by the 3-tuple
K = (p, i, w
1
↑w
2
),
304 Introduction to Computational Complexity
where p is the current state, i is the cell number in the input tape pointed
to by the input head, and w
1
↑w
2
is the contents of the worktape. Here the
head of the worktape is pointing to the ﬁrst symbol of w
2
.
Deﬁnition 11.6 The space used by an oﬀ-line Turing machine M on
input x, denoted by S
M
(x), is deﬁned by:
(1) If there is an accepting computation of M on input x, then S
M
(x)
is the number of worktape cells used in an accepting computation
that uses the least number of worktape cells, and
(2) If there is no accepting computation of M on input x, then S
M
(x) =
∞.
Example 11.2 Consider the following Turing machine M that recognizes the
language L = ¦a
n
b
n
[ n ≥ 1¦. M scans its input tape from left to right and counts
the number of a’s representing the value of this count in binary notation on its
worktape. It does this by incrementing a counter in its worktape. M then veriﬁes
that the number of occurrences of the symbol b is the same by subtracting 1 from
the counter for each b scanned. If n is the length of the input string x, then M
uses ]log(n/2) + 1 worktape cells in order to accept x.
Let L be a language and f a function from the set of nonnegative integers
to the set of nonnegative integers. We say that L is in DSPACE(f) (resp.
NSPACE(f)) if there exists a deterministic (resp. nondeterministic) oﬀ-
line Turing machine M that behaves as follows. On input x, if x ∈ L then
S
M
(x) ≤ f([x[); otherwise S
M
(x) = ∞. For example, L(M) = ¦a
n
b
n
[ n ≥
1¦ in Example 11.2 is in DSPACE(log n) since M is deterministic and for
any string x, if x ∈ L, M uses at most log(n/2) + 1| worktape cells in
order to accept x, where n = [x[. Similarly, we may deﬁne DSPACE(n
k
),
NSPACE(n
k
) for any k ≥ 1. We now deﬁne two important space complexity
classes PSPACE and NSPACE as follows.
PSPACE = DSPACE(n)∪DSPACE(n
2
)∪DSPACE(n
3
)∪. . .∪DSPACE(n
k
)∪. . .
NSPACE = NSPACE(n)∪NSPACE(n
2
)∪NSPACE(n
3
)∪. . .∪NSPACE(n
k
)∪. . .
In other words, PSPACE is the set of all languages recognizable in polyno-
mial space using a deterministic oﬀ-line Turing machine, and NSPACE is
Tape compression and linear speed-up 305
the set of all languages recognizable in polynomial space using a nondeter-
ministic oﬀ-line Turing machine. There are also two fundamental complex-
ity classes:
LOGSPACE = DSPACE (log n) and NLOGSPACE = NSPACE (log n),
which deﬁne the two classes of languages recognizable in logarithmic space
using a deterministic and nondeterministic oﬀ-line Turing machine, respec-
tively. In the following example, we describe a problem that belongs to the
class NLOGSPACE.
Example 11.3 graph accessibility problem (gap): Given a ﬁnite di-
rected graph G = (V, E), where V = ¦1, 2, . . . , n¦, is there a path from vertex 1
to vertex n? Here, 1 is the start vertex and n is the goal vertex. We construct
a nondeterministic Turing machine M that determines whether there is a path
from vertex 1 to vertex n. M performs this task by ﬁrst beginning with the path
of length zero from vertex 1 to itself, and extending the path at each later step
by nondeterministically choosing a next vertex, which is a successor of the last
vertex in the current path. It records in its worktape only the last vertex in this
path; it does not record the entire list of vertices in the path. Since the last vertex
can be represented by writing its number in binary notation on the worktape, M
uses at most ]log(n + 1) worktape cells. Since M chooses a path nondetermin-
istically, if a path from vertex 1 to vertex n exists, then M will be able to make
a correct sequence of choices and construct such a path. It will answer yes when
it detects that the last vertex in the path chosen is n. On the other hand, M is
not forced to make the right sequence of choices, even when an appropriate path
exists. For example, M may loop, by choosing an endless sequence of vertices in
G that form a cycle, or M may terminate without indicating that an appropriate
path exists by making an incorrect choice for the successor of the last vertex in
the path. Since M needs to store in its worktape only the binary representa-
tion of the current vertex whose length is ]log(n + 1), it follows that gap is in
NLOGSPACE.
11.5 Tape compression and linear speed-up
Since the tape alphabet can be arbitrarily large, several tape symbols
can be encoded into one. This results in tape compression by a constant
factor, i.e., the amount of space used is reduced by some constant c > 1.
306 Introduction to Computational Complexity
Similarly, one can speed up the computation by a constant factor. Thus,
in computational complexity, the constant factors may be ignored; only the
rate of growth is important in classifying problems. In the following, we
state without proof two theorems on tape compression and linear speed-up.
Theorem 11.1 If a language L is accepted by an S(n) space-bounded
oﬀ-line Turing machine M, then for any constant c, 0 < c < 1, L is accepted
by a cS(n) space-bounded oﬀ-line Turing machine M

.
Theorem 11.2 If a language L is accepted by a T(n) time-bounded
Turing machine M with k > 1 tapes such that n = o(T(n)), then for
any constant c, 0 < c < 1, L is accepted by a cT(n) time-bounded Turing
machine M

.
Example 11.4 Let L = ¦ww
R
[ w ∈ ¦a, b¦
+
¦, i.e., L consists of the set of
palindromes over the alphabet ¦a, b¦. A 2-tape Turing machine M can be con-
structed to accept L as follows. The input string is initially in the ﬁrst tape. The
second tape is used to mark the input symbols in the following way. Scan the ﬁrst
symbol and mark it. Go to the rightmost symbol, scan it and mark it. Continue
this process until the input string is consumed, in which case it is accepted, or
until a mismatch is found, in which case the input string is rejected. Another
2-tape Turing machine M

that recognizes the language L works as follows. Scan
simultaneously the two leftmost symbols, mark them and go to the right to scan
and mark the two rightmost symbols, etc. Clearly, the time required by M

is
almost half the time required by M.
11.6 Relationships Between Complexity Classes
Deﬁnition 11.7 A total function T from the set of nonnegative integers
to the set of nonnegative integers is said to be time constructible if and only
if there is a Turing machine which on every input of length n halts in exactly
T(n) steps. A total function S from the set of nonnegative integers to the set
of nonnegative integers is said to be space constructible if and only if there is
a Turing machine which on every input of length n halts in a conﬁguration
in which exactly S(n) tape cells of its work space are nonblank, and no
other work space has been used during the computation. Almost all known
functions are time and space constructible, e.g. n
k
, c
n
, n!.
Relationships Between Complexity Classes 307
Theorem 11.3
(a) DTIME(f(n)) ⊆ NTIME(f(n)) and DSPACE(f(n)) ⊆ NSPACE(f(n)).
(b) DTIME(f(n)) ⊆ DSPACE(f(n)) and NTIME(f(n)) ⊆ NSPACE(f(n)).
(c) If S is a space constructible function and S(n) ≥ log n, then
NSPACE(S(n)) ⊆ DTIME(c
S(n)
), c ≥ 2.
Proof.
(a) By deﬁnition, every deterministic Turing machine is nondeterministic.
(b) In n steps, at most n + 1 tape cells can be scanned by the tape heads.
(c) Let M be a nondeterministic oﬀ-line Turing machine such that on all in-
puts of length n, M uses a work space bounded above by S(n) ≥ log n. Let
s and t be the number of states and worktape symbols of M, respectively.
Since M is S(n) space-bounded and S(n) is space constructible, the maxi-
mum number of distinct conﬁgurations that M can possibly enter on input
x of length n is s(n + 2)S(n)t
S(n)
. This is the product of the number of
states, number of input tape head positions, number of worktape head po-
sitions and number of possible worktape contents. Since S(n) ≥ log n, this
expression is bounded above by d
S(n)
for some constant d ≥ 2. Therefore,
M cannot make more than d
S(n)
moves, for otherwise one conﬁguration will
be repeated and the machine will never halt. Without loss of generality, we
may assume that if M accepts, it erases both of its tapes and brings the tape
heads to the ﬁrst cell before entering the accepting state. Consider a deter-
ministic Turing machine M

that, on input x of length n, generates a graph
having all the conﬁgurations of M as its vertices, and setting a directed
edge between two conﬁgurations if and only if the second one is reachable
from the ﬁrst in one step according to the transition function of M. The
number of conﬁgurations is computed using the space constructibility of S.
M

then checks whether there is a directed path in the graph joining the
initial and the unique accepting conﬁguration, and accepts if and only if
this is the case. This can be done in time O(d
2S(n)
) = O(c
S(n)
) for some
constant c ≥ 2 (A shortest path in a directed graph with n vertices can be
found in O(n
2
) time). Obviously, M

can simulate
M using divide and conquer. Let s and t be the number of states and work-
tape symbols of M, respectively. Since M is S(n) space-bounded and S(n)
is space constructible, the maximum number of distinct conﬁgurations that
M can possibly enter on input x of length n is s(n + 2)S(n)t
S(n)
. This is
the product of the number of states, number of input tape head positions,
number of worktape head positions and number of possible worktape con-
tents. Since S(n) ≥ log n, this expression is bounded above by 2
cS(n)
for
some constant c ≥ 1. Therefore, M cannot make more than 2
cS(n)
moves,
for otherwise one conﬁguration will be repeated and the machine will never
halt. Let the initial conﬁguration on input x be C
i
and the ﬁnal conﬁgura-
tion C
f
. M will accept x if and only if x causes the machine to go from C
i
to C
f
. Suppose that this takes M j moves. Then there must exist a conﬁg-
urations C such that x causes M to go into conﬁguration C of size O(S(n))
in j/2 steps and then from C to C
f
in j/2 steps. M

will check for all pos-
sible conﬁgurations C using the divide-and-conquer function reachable
shown below. The ﬁrst call to this function is reachable(C
i
, C
f
, 2
cS(n)
).
1. Function reachable(C
1
, C
2
, j)
2. if j = 1 then
3. if C
1
= C
2
or C
2
is reachable from C
1
in one step
4. then return true
5. else return false
6. end if
7. else for each possible conﬁguration C of size ≤ S(n)
8. if reachable (C
1
, C, j/2) and reachable (C, C
2
, j/2)
9. then return true
10. else return false
11. end if
12. end if
13. end reachable.
Relationships Between Complexity Classes 309
The function reachable decides whether there is a partial computation
of length at most j between two conﬁgurations. It does so by looking for
the middle conﬁguration C and checking recursively that it is indeed the
middle conﬁguration. This checking amounts to verifying the existence of
two partial conﬁgurations of length at most j/2 each.
It is immediately clear that M

accepts its input if and only if M does.
Let us show the space bound for M

. To simulate the recursive calls, M

uses its worktape as a stack, storing in it the information corresponding to
successive calls of the function. Each call decreases the value of j by a factor
of 2. Therefore, the depth of recursion is cS(n), and hence no more than
cS(n) calls are active simultaneously. For each call, M

) = L(M). ⁄
Corollary 11.2 For any k ≥ 1,
NSPACE(n
k
) ⊆ DSPACE(n
2k
) and NSPACE(log
k
n) ⊆ DSPACE(log
2k
n).
Moreover, NSPACE = PSPACE.
Corollary 11.3 There is a deterministic algorithm to solve the problem
gap using O(log
2
n) space.
Proof. Immediate from Theorem 11.4 and the fact that gap has a non-
deterministic algorithm that uses O(log n) space (see Example 11.3). ⁄
11.6.1 Space and time hierarchy theorems
Now we present two hierarchy theorems which are concerned with the re-
lationships between classes when the same resource on the same model is
bounded by diﬀerent functions. Speciﬁcally, we will present some suﬃcient
conditions for the strict inclusion between deterministic time and space
classes. These theorems are known as the space hierarchy and time hier-
archy theorems. Let M be a 1-tape Turing machine. We encode M as a
string of 0’s and 1’s corresponding to a binary number as follows. Assume
without loss of generality that the input alphabet of M is ¦0, 1¦, and the
blank is the only additional tape symbol. For convenience, call the symbols
0, 1 and the blank X
1
, X
2
and X
3
, respectively, and denote by D
1
, D
2
and
D
3
the directions L, R and P. Then a move δ(q
i
, X
j
) = (q
k
, X
l
, D
m
) is
310 Introduction to Computational Complexity
encoded by the binary string 0
i
10
j
10
k
10
l
10
m
. Thus, the binary code for
M is 111C
1
11C
2
11 . . . 11C
r
111, where each C
i
is the code for one move
as shown above. Each Turing machine may have many encodings, as the
encodings of moves can be listed in any order. On the other hand, there
are binary numbers that do not correspond to any encodings of Turing ma-
chines. These binary numbers may collectively be taken as the encodings
of the null Turing machine, i.e., the Turing machine with no moves. It fol-
lows that we may talk of the nth Turing machine, and so on. In a similar
manner, we can encode k-tape Turing machines for all k ≥ 1, and oﬀ-line
Turing machines.
Theorem 11.5 Let S(n) and S

(n) be two space constructible space
bounds, and assume that S

(n) is o(S(n)). Then, DSPACE(S(n)) contains
a language that is not in DSPACE(S

(n)).
Proof. The proof is by diagonalization. Without loss of generality, we
may consider only oﬀ-line Turing machines with input alphabet ¦0, 1¦. We
may also assume that a preﬁx of any number of 1’s is permitted in any
encoding of a Turing machine, so that each Turing machine has inﬁnitely
many encodings. We construct a Turing machine M with space bound S(n)
that disagrees on at least one input with any Turing machine with space
bound S

(n). M treats its input x as an encoding of an oﬀ-line Turing
machine. Let x be an input to M of length n. First, to ensure that M
does not use more than S(n) space, it ﬁrst marks exactly S(n) cells of its
worktape. Since S(n) is space constructible, this can be done by simulating
a Turing machine that uses exactly S(n) space on each input of length n.
From now on, M aborts its operation whenever the computation attempts
to use a cell beyond the marked cells. Thus, M is indeed an S(n) bounded
Turing machine, that is, L(M) is in DSPACE(S(n)). Next, M simulates
M
x
on input x, where M
x
is the Turing machine whose encoding is the
input x. M accepts x if and only if it completes the simulation using S(n)
space and M
x
halts and rejects x. If M
x
is S

(n).
It should be noted that L(M) may be accepted by a Turing machine
other than M. We now show that if a Turing machine M

accepts L(M),
then M

cannot be S

(n) space bounded. For suppose that there exists an
S

(n) space bounded Turing machine M

that accepts L(M), and assume
without loss of generality that M

halts on all inputs. Since S

(n) is o(S(n)),
Relationships Between Complexity Classes 311
and since any oﬀ-line Turing machine can have an encoding with arbitrarily
many 1’s, there exists an encoding x

of M

such that log t|S

(n

) < S(n

),
where n

= [x

[. Clearly, on input x

, M has suﬃcient space to simulate M

.
But then, on input x

, M will accept if and only if M

halts and rejects.
It follows that L(M

) = L(M), and hence L(M) is in DSPACE(S(n)) and
not in DSPACE(S

(n)). ⁄
For the time hierarchy theorem, we need the following lemma whose
proof is omitted:
Lemma 11.1 If L is accepted by a k-tape Turing machine in time T(n),
then L is accepted by a 2-tape Turing machine in time T(n) log T(n).
Theorem 11.6 Let T(n) and T

(n) be two time bounds such that T(n)
is time constructible and T

(n) log T

(n) is o(T(n)). Then, DTIME(T(n))
contains a language which is not in DTIME(T

(n)).
Proof. The proof is similar to that of Theorem 11.5. Therefore, we will
only state here the necessary modiﬁcations. On input x of length n, M
shuts itself oﬀ after executing exactly T(n) steps. This can be done by
simulating a T(n) time bounded Turing machine on extra tapes (note that
this is possible since T(n) is time constructible). It should be noted that
M has only a ﬁxed number of tapes, and it is supposed to simulate Turing
machines with arbitrarily many tapes. By Lemma 11.1, this results in a
slowdown by a factor of log T

(n). Also, as in the proof of Theorem 11.5,
M

may have many tape symbols, which slows down the simulation by
a factor of c = log t|, where t is the number of tape symbols used by
M

. Thus, the encoding x

of M

of length n

must satisfy the inequality
cT

(n

) log T

(n

) ≤ T(n

). Since M accepts x

only if M

halts and rejects
x

, it follows that L(M

) = L(M), and hence L(M) is in DTIME(T(n))
and not in DTIME(T

(n)). ⁄
11.6.2 Padding arguments
Suppose we are given any particular problem Π. Then, we can create a
version of Π that has lower complexity by padding each instance of Π with
a long sequence of extra symbols. This technique is called padding. We
illustrate the idea behind this concept in connection with an example. Let
L ⊆ Σ
∗
be a language, where Σ is an alphabet that does not contain the
312 Introduction to Computational Complexity
symbol 0. Suppose that L is in DTIME(n
2
). Deﬁne the language
L

= ¦x0
k
[ x ∈ L and k = [x[
2
−[x[¦.
L

is called a padded version of L. Now we show that L

is in DTIME(n).
Let M be a Turing machine that accepts L. We construct another Turing
machine M

that recognizes L

as follows. M

ﬁrst checks that the input
string x

is of the form x0
k
, where x ∈ Σ
∗
and k = [x[
2
−[x[. This can be
done in an amount of time bounded by [x

[. Next, if x

is of the form x0
k
,
then M

simulates on input x

= x0
k
the computation of M on input x.
If M accepts x, then M

accepts; otherwise M

rejects. Since M requires
at most [x[
2
steps to decide if x is in the language L, M

needs at most
[x

[ = [x[
2
steps to decide if x

is in the language L

. Therefore, L

is
in DTIME(n). In more general terms, if L is in DTIME(f(n
2
)), then L

is in DTIME(f(n)). For example, if L is in DTIME(n
4
), then L

is in
DTIME(n
2
), and if L is in DTIME(2
n
2
), then L

is in DTIME(2
n
).
We now present two theorems that are based on padding arguments.
Theorem 11.7 If DSPACE(n) ⊆ P, then PSPACE = P.
Proof. Assume that DSPACE(n) ⊆ P. Let L ⊆ Σ
∗
be a set in PSPACE,
where Σ is an alphabet that does not contain the symbol 0. Let M be
a Turing machine that accepts L in space p(n) for some polynomial p.
Consider the set
L

in polynomial
time. Clearly, another deterministic Turing machine, which on input x
appends to it 0
k
, where k = 2
cn
−[x[, and then simulates M

can easily be
constructed. Obviously, this machine accepts L in time 2
cn
. It follows that
NTIME(2
cn
) ⊆ DTIME(2
cn
). Since DTIME(2
cn
) ⊆ NTIME(2
cn
), we have
as a result NTIME(2
cn
) = DTIME(2
cn
). Since c is arbitrary, it follows that
NEXT = DEXT. ⁄
In other words, the above theorem says that if NEXT = DEXT, then
there is a language L that is recognizable in linear time by a nondetermin-
istic Turing machine, but not recognizable by any polynomial time deter-
ministic Turing machine.
Corollary 11.5 If NP = P, then NEXT = DEXT.
11.7 Reductions
In this section we develop methods for comparing complexity classes of
computational problems. Such comparisons will be made by describing
transformations from one problem to another. A transformation is simply
a function that maps instances of one problem into instances of another
problem. Let A ∈ Σ
∗
and B ∈ ∆
∗
be two arbitrary problems, which are
encoded as sets of strings over the alphabets Σ and ∆, respectively. A func-
tion f which maps strings over the alphabet Σ into strings over the alphabet
∆ is a transformation of A into B, if the following property is satisﬁed:
∀x ∈ Σ
∗
x ∈ A if and only if f(x) ∈ B.
A transformation f from A to B is useful, since it implies a transfor-
mation also from any algorithm to solve B into an algorithm to solve A.
314 Introduction to Computational Complexity
That is, one may construct the following algorithm to solve the problem A,
given as input an arbitrary string x ∈ Σ
∗
:
(1) Transform x into f(x).
(2) Decide whether f(x) ∈ B or not.
(3) If f(x) ∈ B, then answer yes; otherwise answer no.
The complexity of this algorithm to solve A depends upon two factors:
the complexity of transforming x into f(x), and the complexity of deciding
whether a given string is in B or not. However, it is clear that an eﬃcient
algorithm for B will be transformed into an eﬃcient algorithm for A by the
above process if the transformation is not too complex.
Deﬁnition 11.8 If there is a transformation f from a problem A to a
problem B, then we say that A is reducible to B, denoted by A ∝ B.
Deﬁnition 11.9 Let A ⊆ Σ
∗
and B ⊆ ∆
∗
be sets of strings. Suppose
that there is a transformation f : Σ
∗
→ ∆
∗
. Then
• A is polynomial time reducible to B, denoted by A ∝
poly
B, if f(x)
can be computed in polynomial time.
• A is log space reducible to B, denoted by A ∝
log
B, if f(x) can be
computed using O(log [x[) space.
Deﬁnition 11.10 Let ∝ be a reducibility relation. Let L be a family of
languages. Deﬁne the closure of L under the reducibility relation ∝ by
closure
∝
(L) = ¦L [ ∃L

∈ L (L ∝ L

)¦.
Then, L is closed under the reducibility relation ∝ if and only if
closure
∝
(L) ⊆ L.
If L consists of one language L, then we will write closure
∝
(L) instead of
closure
∝
(¦L¦).
For example, closure
∝
poly
(P) is the set of all languages that are re-
ducible to P in polynomial time, and closure
∝
log
(P) is the set of all lan-
guages that are reducible to P in log space. We will show later that P
is closed under both the reducibility relations ∝
poly
and ∝
log
by showing
that closure
∝
poly
(P) ⊆ P and closure
∝
log
(P) ⊆ P.
Now, we establish the relationship between the two important forms of
reducibility: polynomial time and log space reducibilities.
Reductions 315
Lemma 11.2 The number of distinct conﬁgurations that a log space
bounded oﬀ-line Turing machine M can enter with an input of length n is
bounded above by a polynomial in n.
Proof. Let s and t be the number of states and worktape symbols of M,
respectively. The number of distinct conﬁgurations that M can possibly
enter on an input of length n is given by the product of the following quan-
tities: s (the number of states of M), n + 2 (the number of distinct input
head positions of M on an input of length n plus the left and right mark-
ers), log n (the number of distinct worktape head positions), and t
log n
(the
number of distinct strings that can be written within the log n worktape
cells). Thus, the number of distinct conﬁgurations of M on an input of
length n is
s(n + 2)(log n)t
log n
= s(n + 2)(log n)n
log t
≤ n
c
, c > 1,
for all but ﬁnitely many n. It follows that the number of conﬁgurations is
bounded by a polynomial in n. ⁄
Theorem 11.9 For any two languages A and B,
if A ∝
log
B then A ∝
poly
B.
Proof. Immediate from Lemma 11.2 (also Corollary 11.1). ⁄
Consequently, any log space reduction is a polynomial time reduction.
It follows that for any family L of languages, if L is closed under polynomial
time reductions, then it is also closed under log space reductions.
Lemma 11.3 P is closed under polynomial time reductions.
Proof. Let L ⊆ Σ
∗
, for some ﬁnite alphabet Σ, be any language such that
L ∝
poly
L

for some language L

∈ P. By deﬁnition, there is a function f
computable in polynomial time, such that
∀x ∈ Σ
∗
x ∈ L if and only if f(x) ∈ L

.
Since L

∈ P, there exists a deterministic Turing machine M

that accepts
L

and operates in time n
k
, for some k ≥ 1. Since f is computable in poly-
nomial time, there exists a deterministic Turing machine M

that computes
f and operates in time n
l
, for some l ≥ 1. We construct a Turing machine
316 Introduction to Computational Complexity
M that accepts the set L. M performs the following steps on input x over
the alphabet Σ:
(1) Transform x into f(x) using Turing machine M

.
(2) Determine whether f(x) ∈ L

or not using the Turing machine M

.
(3) If M

decides that f(x) ∈ L

, then accept; otherwise, do not accept.
The time complexity of this algorithm for the Turing machine M is simply
the sum of the amounts of time spent doing Steps (1), (2), and (3). Let x be
a string of length n and let f(x) be a string of length m. Then, the amount
of time used by this algorithm on input x is bounded by n
l
+m
k
+1, since
Step (1) takes at most n
l
, Step (2) at most m
k
steps, and Step (3) one
step. We observe that f(x) cannot be longer than n
l
, since M

operates in
n
l
steps and at most one symbol is printed by M

on the output tape per
step. In other words, m ≤ n
l
. Therefore, n
l
+m
k
+1 ≤ n
l
+n
kl
+1, for all
but ﬁnitely many n. We have demonstrated thereby that a deterministic
Turing machine exists that recognizes the set L in polynomial time. Thus,
if L ∝
poly
L

then L ∈ P. ⁄
The proof of the following lemma is similar to the proof of Lemma 11.3:
Lemma 11.4 NP and PSPACE are closed under polynomial time reduc-
tions.
Corollary 11.6 P, NP and PSPACE are closed under log space reduc-
tions.
Lemma 11.5 LOGSPACE is closed under log space reductions.
Proof. Let L ⊆ Σ
∗
, for some ﬁnite alphabet Σ, be any language such
that L ∝
log
L

for some language L

∈ LOGSPACE. By deﬁnition, there is
a function f computable in log space such that
∀x ∈ Σ
∗
x ∈ L if and only if f(x) ∈ L

.
Since L

∈ LOGSPACE, there exists a deterministic Turing machine M

that accepts L

in space log n. Since f is computable in log space, there
exists a deterministic Turing machine M

that computes f using at most
log n worktape cells on input of size n. We construct a deterministic Turing
machine M that accepts the set L. M performs the following steps on input
x over the alphabet Σ:
Reductions 317
(1) Set i to 1.
(2) If 1 ≤ i ≤ [f(x)[, then compute the ith symbol of f(x) using the Turing
machine M

. Call this symbol σ. If i = 0, then let σ be the left endmarker
symbol#. If i = [f(x)[ + 1, then let σ be the right endmarker symbol $.
(3) Simulate the actions of the Turing machine M

on the symbol σ until
the input head of M

moves right or left. If the input head moves to the
right, then add one to i and go to Step (2). If the input head of M

moves
to the left, then subtract one from i and go to Step (2). If M

enters a ﬁnal
state before moving its input head either right or left, thereby accepting
f(x), then accept the input string x.
It should be noted that M does indeed recognize the set L. It accepts
a string x if and only if M

works in log n space. The worktape
contents of the simulated Turing machine M

are stored in the worktape
space of M. This needs at most log [f(x)[ space, since M

is a log space
Turing machine and it is being simulated on input f(x). As we have seen
in Lemma 11.2, M

is polynomially time bounded, since M

operates in
space log n and eventually terminates with the value of the function f.
Therefore, [f(x)[ ≤ [x[
c
, for some c > 0. Thus, the worktape space needed
for representing the contents of the worktape of M

can be stored on the worktape of M using binary notation
within space log [f(x)[ ≤ log [x[
c
= c log [x[ worktape cells. Therefore, the
algorithm described for the Turing machine M requires at most d log n
worktape cells, for some d > 0, to recognize the set L. It follows that L is
in LOGSPACE and hence closure
∝
log
(LOGSPACE) ⊆ LOGSPACE. ⁄
The following lemma is proven in the same manner as Lemma 11.5.
Lemma 11.6 NLOGSPACE is closed under log space reductions.
318 Introduction to Computational Complexity
11.8 Completeness
Deﬁnition 11.11 Let ∝ be a reducibility relation, and L a family of
languages. A language L is complete for L with respect to the reducibility
relation ∝ if L is in the class L and every language in L is reducible to the
language L by the relation ∝, that is, L ⊆ closure
∝
(L).
We have presented in Chapter 10 some problems that are complete for
the class NP with respect to polynomial time reductions. In fact, most of
the reductions in the proofs of NP-completeness found in the literature are
log space reductions.
We observe that every set S ∈ LOGSPACE is log space reducible to a
set with just one element. That is, given a set S ⊆ Σ
∗
in LOGSPACE, we
deﬁne the function f
S
by
f
S
(x) =

1 if x ∈ S
0 otherwise.
It follows, trivially, that the set ¦1¦ is LOGSPACE-complete with respect to
log space reduction. In fact, every problem in LOGSPACE is LOGSPACE-
complete with respect to log space reduction. This is because log space
reductions are too powerful to distinguish between sets in LOGSPACE.
11.8.1 NLOGSPACE-complete problems
In the following theorem, we prove that the problem gap is complete for
the class NLOGSPACE.
Theorem 11.10 gap is log space complete for the class NLOGSPACE.
Proof. We have shown in Example 11.3 that gap is in the class
NLOGSPACE. It remains to show that any problem in that class reduces
to gap using log space reduction. Let L be in NLOGSPACE. We show
that L ∝
log
gap. Since L is in NLOGSPACE, there is a nondeterministic
oﬀ-line Turing machine M that accepts L, and for every x in L, there is an
accepting computation by M that visits at most log n worktape cells, where
n = [x[. We construct a log space reduction which transforms each input
string x into an instance of the problem gap consisting of a directed graph
G = (V, E). The set of vertices V consists of the set of all conﬁgurations
Completeness 319
K = (p, i, w
1
↑ w
2
) of M on input x such that [w
1
w
2
[ ≤ log n. The set of
edges consists of the set of pairs (K
1
, K
2
) such that M can move in one
step on input x from the conﬁguration K
1
to the conﬁguration K
2
. Further-
more, the start vertex s is chosen to be the initial conﬁguration K
i
of M on
input x. If we assume that when M enters the ﬁnal state q
f
, it erases all
symbols in its worktape and positions its input head on the ﬁrst cell, then
the goal vertex t is chosen to be the ﬁnal conﬁguration K
f
= (p
f
, 1, ↑B).
It is not hard to see that M accepts x within log n worktape space if and
only if G has a path from its start vertex s to its goal vertex t. To ﬁnish
the proof, note that G can be constructed using only O(log n) space. ⁄
Corollary 11.7 gap is in LOGSPACE if and only if
NLOGSPACE = LOGSPACE.
Proof. If NLOGSPACE = LOGSPACE, then clearly gap is in
LOGSPACE. On the other hand, assume that gap is in LOGSPACE.
Then,
closure
∝
log
(gap) ⊆ closure
∝
log
(LOGSPACE) ⊆ LOGSPACE,
since LOGSPACE is closed under ∝
log
. Since gap is complete for the class
NLOGSPACE, we have
NLOGSPACE ⊆ closure
∝
log
(gap).
Thus NLOGSPACE ⊆ LOGSPACE. Since LOGSPACE ⊆ NLOGSPACE,
it follows that NLOGSPACE = LOGSPACE. ⁄
The proof of the following theorem is left as an exercise (Exercise 11.27).
Theorem 11.11 2-SAT is log space complete for the class NLOGSPACE.
11.8.2 PSPACE-complete problems
Deﬁnition 11.12 A problem Π is PSPACE-complete if it is in PSPACE
and all problems in PSPACE can be reduced to Π using polynomial time
reduction.
320 Introduction to Computational Complexity
The relationship of the following problem to PSPACE is similar to the
relationship of the problem satisfiability to NP.
quantified boolean formulas (qbf): Given a boolean expression E on
n variables x
1
, x
2
, . . . , x
n
, is the boolean formula
F = (Q
1
x
1
)(Q
2
x
2
) . . . (Q
n
x
n
)E
true? Here each Q
i
is either ∃ or ∀.
Theorem 11.12 quantified boolean formula is PSPACE-
complete.
That quantified boolean formula is in PSPACE follows from the
fact that we can check whether F is true by trying all the possible truth
assignments for the variables x
1
, x
2
, . . . , x
n
and evaluating E for each. It is
not hard to see that no more than polynomial space is needed, even though
exponential time will be required to examine all 2
n
truth assignments. The
proof that each language L ∈ PSPACE can be transformed to quantified
boolean formula is similar to the proof that the problem satisfiability
is NP-complete.
An interesting PSPACE-complete problem is the following:
csg recognition: Given a context-sensitive grammar G and a string x,
is x ∈ L(G)? Here L(G) is the language generated by G.
It is well known that the class NSPACE(n) is precisely the set of languages
generated by context-sensitive grammars. This problem can be rephrased
in terms of Turing machines as follows. A linear bounded automaton is a
restricted type of Turing machine in which the worktape space consists of
n + 2 cells, where n is the input length. Thus, equivalently, the following
problem is PSPACE-complete:
lba acceptance: Given a nondeterministic linear bounded automaton M
and a string x, does M accept x?
This problem remains PSPACE-complete even if the Turing machine is
deterministic. Thus, all problems that are solvable in polynomial space can
be reduced in polynomial time to a problem that requires only linear space.
In addition to the above problems, the set of PSPACE-complete prob-
lems includes many interesting problems in widely diﬀerent areas, espe-
cially in game theory. Several two-person games involving a natural alter-
nation of turns for the two players which correspond to an alternation of
quantiﬁers in quantified boolean formula are known to be PSPACE-
Completeness 321
complete. For example, generalized versions of the games hex, geography
and kayles are PSPACE-complete. Also generalized versions of the more
familiar games checkers and go are known to be PSPACE-complete un-
der certain drawing restrictions.
11.8.3 P-complete problems
Although the class P contains all problems for which there exists an eﬃcient
algorithm, there are problems in P that are practically intractable. The
following example reveals the hardness of a practical problem in this class.
Example 11.5 Consider the problem k-clique deﬁned as follows. Given an
undirected graph G = (V, E) with n vertices, determine whether G contains a
clique of size k, where k is ﬁxed. The only known algorithm to solve this problem
is by considering all the k subsets of V . This results in Ω(n
k
/k!) time complexity.
Thus, even for moderate values of k, the problem is practically intractable.
Deﬁnition 11.13 A problem Π is P-complete if it is in P and all problems
in P can be reduced to Π using log space reduction.
It is widely conjectured that there are problems in P for which any
algorithm must use an amount of space that is more than logarithmic in
the input size, that is, the set P −LOGSPACE is not empty.
The class of P-complete problems is not empty, and it does contain
several problems that are solvable in polynomial time of low degree such
as depth-ﬁrst search, which is solvable in linear time, and the max-ﬂow
problem, which is solvable in O(n
3
) time. These problems are important
in the ﬁeld of parallel algorithms, as they contain those problems which
are hard to parallelize eﬃciently; they usually admit sequential algorithms
that are greedy in nature and thus inherently sequential.
Deﬁnition 11.14 The class NC consists of those problems that can be
solved in polylogarithmic time, that is O(log
k
n) time, using a polynomial
number of processors.
This class remains invariant under diﬀerent models of parallel computa-
tion. It encompasses those problems that are well-parallelizable in the sense
that increasing the number of processors results in signiﬁcant speedup. Ob-
serve that NC ⊆ P, as the total number of steps performed by a parallel
algorithm is the product of the running time and the number of processors,
322 Introduction to Computational Complexity
which is polynomial in the case of NC algorithms. In other words, such
a parallel algorithm can be transformed into a polynomial time sequential
algorithm.
However, there is a general belief that NC = P. Interestingly, if a
problem is P-complete, then every other problem in P can be reduced to
it in polylogarithmic time using a polynomial number of processors. This
type of transformation is called NC-reduction. It can be shown that NC is
closed under NC-reduction. This motivates the next alternative deﬁnition
of P-complete problems.
Deﬁnition 11.15 A problem Π is P-complete if it is in P and all problems
in P can be reduced to Π using NC-reduction.
This deﬁnition yields the following theorem:
Theorem 11.13 If a problem Π is P-complete and Π is in NC, then P =
NC.
In other words, if P = NC, then all P-complete problems must be-
long to P − NC. Thus, although P-complete problems are not likely to
be solvable in logarithmic space, they also do not seem to admit eﬃcient
parallel algorithms. The following problem is the ﬁrst problem identiﬁed
as P-complete.
path system accessibility: Given a ﬁnite set V of vertices, a relation
R ⊆ V V V , and two sets S, T ⊆ V of “source” and “terminal” vertices,
is there a smallest subset U ⊆ T such that S ⊆ U and for any two vertices
y, z ∈ U, if (x, y, z) ∈ R then x ∈ U?
It is not hard to see that path system accessibility∈ P (Exercise 11.40)
The following is a sample of other P-complete problems.
(1) circuit value problem(cvp): Given a boolean circuit C consist-
ing of m gates ¦g
1
, g
2
, . . . , g
m
¦, and a speciﬁed set of input val-
ues ¦x
1
, x
2
, . . . , x
n
¦, determine whether the output of the circuit is
equal to 1. Here a gate is ∨, ∧ or .
(2) ordered depth-first search: Given a directed graph G =
(V, E) and three vertices s, u, v ∈ V , determine whether u is visited
before v in a depth-ﬁrst search traversal of G starting at s.
(3) linear programming: Given an n m matrix A of integers, a
vector b of n integers, a vector c of m integers, and an integer k,
Completeness 323
determine whether there exists a vector x of m nonnegative rational
numbers such that Ax ≤ b and cx ≥ k.
(4) max-flow: Given a weighted directed graph G = (V, E) with two
distinguished vertices s and t, determine whether the maximum
ﬂow from s to t is odd.
11.8.4 Some conclusions of completeness
Theorem 11.14 Let Π be an NP-complete problem with respect to poly-
nomial time reductions. Then NP = P if and only if Π ∈ P.
Proof. The theorem is easily established using the deﬁnition of com-
pleteness. Suppose that NP = P. Since Π is complete for NP, Π ∈ NP,
and hence Π ∈ P. On the other hand, suppose that Π ∈ P. Since Π is
NP-complete, NP ⊆ closure
∝
poly
(Π). Thus,
NP ⊆ closure
∝
poly
(Π) ⊆ closure
∝
poly
(P) ⊆ P,
as P is closed under ∝
poly
. Since P ⊆ NP, it follows that NP = P. ⁄
Theorem 11.14 is also true when the problem Π is complete for NP
with respect to log space reductions. This results in the following stronger
theorem, whose proof is similar to the proof of Theorem 11.14.
Theorem 11.15 Let Π be a problem that is complete for NP with respect
to log space reductions. Then
(1) NP = P if and only if Π ∈ P.
(2) NP = NLOGSPACE if and only if Π ∈ NLOGSPACE.
(3) NP = LOGSPACE if and only if Π ∈ LOGSPACE.
In comparing Theorem 11.14 with Theorem 11.15, the number of con-
clusions that can be drawn from knowing that a problem Π is log space
complete for the class NP is more than the number of conclusions that can
be drawn from knowing that Π is complete for NP with respect to polyno-
mial time reductions. In fact, most, if not all, polynomial time reductions
between natural NP-complete problems described in the literature are also
log space reductions. Also, log space reductions can distinguish between
the complexity of sets in P and polynomial time reductions cannot. The
proofs of the following theorems are similar to the proof of Theorems 11.14.
324 Introduction to Computational Complexity
Theorem 11.16 Let Π be a problem that is complete for the class
PSPACE with respect to log space reductions. Then
(1) PSPACE = NP if and only if Π ∈ NP.
(2) PSPACE = P if and only if Π ∈ P.
Theorem 11.17 If a problem Π is P-complete, then
(1) P = LOGSPACE if and only if Π is in LOGSPACE.
(2) P = NLOGSPACE if and only if Π is in NLOGSPACE.
The following theorem is a generalization of Corollary 11.7.
Theorem 11.18 Let Π be a problem that is complete for NLOGSPACE
with respect to log space reductions. Then
NLOGSPACE = LOGSPACE if and only if Π ∈ LOGSPACE.
11.9 The Polynomial Time Hierarchy
An oracle Turing machine is a k-tape Turing machine with an additional
tape called the oracle tape, and a special state called the query state. The
purpose of the oracle is to answer questions about the membership of an
element in an arbitrary set. Let M be a Turing machine for an arbitrary
set A with an oracle for another arbitrary set B. Whenever M wants to
know whether an element x is in the set B, it writes x on its oracle tape,
and then enters its query state. The oracle answers this question in one
step: It erases the oracle tape and then prints yes on the oracle tape if the
string x is in the set B and no if the string x is not in the set B. M can
consult the oracle more than once. Thus, it may ask during the course of
a computation whether each of the strings x
1
, x
2
, . . . , x
k
are in the set B.
Let A and B be arbitrary sets. A is said to be recognizable determin-
istically (nondeterministically) in polynomial time using an oracle for B
if there is a deterministic (nondeterministic) oracle Turing machine which
accepts the set A using an oracle for B and, for some ﬁxed k ≥ 1, takes at
most [x[
k
steps on any input string x.
The Polynomial Time Hierarchy 325
Deﬁnition 11.16 If a language A is accepted by a deterministic oracle
Turing machine in polynomial time using an oracle for the language B, then
A is said to be polynomial time Turing reducible to B.
Let P
B
denote the family of all languages recognizable deterministically
in polynomial time using an oracle for the set B, and let NP
B
denote the
family of all languages recognizable nondeterministically in polynomial time
using an oracle for the set B. Let T be a family of languages. The family
co-T denotes the family of complements of sets in T. That is, co-T = ¦co-
S [ S ∈ T¦.
Deﬁnition 11.17 The polynomial time hierarchy consists of the families
of sets ∆
p
i
, Σ
p
i
, Π
p
i
, for all integers i ≥ 0, deﬁned by
∆
p
0
= Σ
p
0
= Π
p
0
= P,
and for all i ≥ 0

∆
p
i+1
=
¸
B∈Σ
p
i
P
B
Σ
p
i+1
=
¸
B∈Σ
p
i
NP
B
Π
p
i+1
= co-Σ
p
i+1
.
The following theorems summarize some of the properties of the classes
in the polynomial time hierarchy. In these theorems, we will use the more
general concept of algorithm in place of Turing machines.
Theorem 11.19 ∆
p
1
= P, Σ
p
1
= NP and Π
p
1
= co-NP.
Proof. We show that, for any set B in P, every set A in P
B
is again
in P. Let the oracle set B be recognized in polynomial time by a deter-
ministic algorithm T
B
that runs in cn
k
steps, for some c > 0. Let T
A
be an algorithm that accepts the set A in polynomial time using oracle B
and runs in dn
l
steps, for some d > 0. One can replace each request for
an answer from the oracle in T
A
by the execution of the algorithm T
B
,
which decides the membership in the set B. Since the algorithm T
A
runs
in dn
l
steps, the maximum length of any question, i.e., string, to the ora-
cle is dn
l
. In replacing each one of these requests by an execution of the
algorithm T
B
, we make each such step of T
A
take at most c(dn
l
)
k
steps.
326 Introduction to Computational Complexity
So, the new algorithm recognizes A without the use of an oracle in at most
dn
l
(cd
k
n
kl
) steps. Since dn
l
cd
k
n
kl
≤ c

n
kl+l
, for some constant c

> 0, it
follows that there is a polynomial time algorithm for A. Thus, for every set
B ∈ Σ
p
0
= P, P
B
⊆ P. It follows that ∆
p
1
=
¸
B∈Σ
p
0
P
B
⊆ P. To ﬁnish the
proof, note that P = P
φ
and the empty set φ is in Σ
p
0
, that is P ⊆ ∆
p
1
. The
proof that Σ
p
1
= NP is similar. It follows, by deﬁnition, that Π
p
1
= co-NP.
⁄
Theorem 11.20 For all i ≥ 0, Σ
p
i
∪ Π
p
i
⊆ ∆
p
i+1
.
Proof. By deﬁnition, ∆
p
i+1
=
¸
B∈Σ
p
i
P
B
. Since a polynomial time algo-
rithm can easily be constructed to accept B using the set B as an oracle, it
follows that Σ
p
i
⊆ ∆
p
i+1
. Also, as we have seen, a polynomial time algorithm
can easily be constructed to accept co-B using an oracle for B. Therefore,
Π
p
i
= co−Σ
p
i
⊆ ∆
p
i+1
. ⁄
Theorem 11.21 For all i ≥ 1, ∆
p
i
⊆ Σ
p
i
∩ Π
p
i
.
Proof. First, ∆
p
i
=
¸
B∈Σ
p
i−1
P
B
⊆
¸
B∈Σ
p
i−1
NP
B
= Σ
p
i
, since a nonde-
terministic algorithm is a generalization of a deterministic algorithm. To
show that ∆
p
i
⊆ Π
p
i
, for all i ≥ 1, it is suﬃcient to show that co-∆
p
i
= ∆
p
i
.
That is, since ∆
p
i
⊆ Σ
p
i
, we have also that co-∆
p
i
⊆ co-Σ
p
i
= Π
p
i
. Thus, if
∆
p
i
= co-∆
p
i
, then ∆
p
i
⊆ Π
p
i
. So, we must show that co-∆
p
i
= ∆
p
i
. Let A be
a set in ∆
p
i
=
¸
B∈Σ
p
i−1
P
B
. Then, there is a deterministic polynomial time
algorithm M
A
for accepting A which uses an oracle for a set B in the family
Σ
p
i−1
. An algorithm M

A
for co-A can be constructed which uses an oracle
for B. M

A
simply stops and accepts if M
A
does not accept, and stops and
rejects if M
A
does stop and accept. It follows that co-∆
p
i
= ∆
p
i
. ⁄
The known relationships between the classes in the polynomial time
hierarchy are shown in Fig. 11.1.
Theorem 11.22 If Σ
p
i
= Σ
p
i+1
for some i ≥ 0, then Σ
p
i+j
= Σ
p
i
for all
j ≥ 1.
Proof. We prove this by induction on j. Assume Σ
p
i+j
= Σ
p
i
, for some
j ≥ 1. Then,
Σ
p
i+j+1
=
¸
B∈Σ
p
i+j
NP
B
=
¸
B∈Σ
p
i
NP
B
= Σ
p
i+1
= Σ
p
i
.
So, Σ
p
i+j+1
= Σ
p
i
, and hence Σ
p
i+j
= Σ
p
i
for all j ≥ 1. ⁄
The Polynomial Time Hierarchy 327
Fig. 11.1 Inclusion relations among complexity classes in the polynomial time hierarchy.
.
Corollary 11.8 If NP = P, then the polynomial time hierarchy collapses,
that is, each family in the polynomial time hierarchy coincides with P.
In fact, the converse implication is also true. If, for any i ≥ 1, it were
true that Σ
p
i
= P, then since NP ⊆ Σ
p
1
⊆ Σ
p
i
, it would follow that NP = P.
Thus, any set A that is complete for the class Σ
p
i
, for any i ≥ 1, satisﬁes
the property that it is in P if and only if NP = P.
Many problems which are not known to be in NP are in the polyno-
mial time hierarchy. Several of these problems are related to NP-complete
problems, but are concerned with ﬁnding a maximum or a minimum. The
following are two examples of problems in P
NP
and NP
NP
.
Example 11.6 chromatic number. Given an undirected graph G = (V, E)
and a positive integer k, is k the smallest number of colors that can be assigned to
the vertices of G such that no two adjacent vertices are assigned the same color?
Recall that the problem coloring stated on page 280 is the problem of deciding
whether it is possible to color a given graph using k colors, where k is a positive
number that is part of the input. It is well known that the problem coloring
is NP-complete. An algorithm to accept chromatic number using an oracle for
coloring is as follows.
(1) If (G, k) is not in coloring then stop and reject, otherwise continue.
(2) If (G, k −1) is in coloring then stop and reject, otherwise continue.
328 Introduction to Computational Complexity
(3) Stop and accept.
We observe that checking whether (G, k) is in coloring is implemented
by asking the oracle coloring and is answered in one step, by assump-
tion. So, the algorithm presented above is clearly polynomial time bounded,
since it needs at most two steps to either accept or reject. It follows that
chromatic number is in ∆
p
2
= P
NP
.
Example 11.7 minimum equivalent expression. Given a well-formed
boolean expression E and a nonnegative integer k, is there a well-formed boolean
expression E

that contains k or fewer occurrences of literals such that E

is
equivalent to E (i.e. E

if and only if E)?
minimum equivalent expression does not appear to be in ∆
p
2
. It is not ob-
vious whether an oracle for a problem in NP can be used to solve minimum equiv-
alent expression in deterministic polynomial time. However, this problem can
be solved in nondeterministic polynomial time using an oracle for satisfiability.
The algorithm is as follows
(1) Guess a boolean expression E

)) is
satisﬁable.
(3) If it is not satisﬁable then stop and accept, otherwise stop and reject.
The correctness of the above algorithm follows from the fact that a well-formed
formula E is not satisﬁable if and only if its negation is a tautology. Thus, since
we want (E

if and only if E) to be a tautology, we only need to check whether
((E

→ E) ∧ (E → E

))
is not satisﬁable. As to the time needed, Step 1, generating E

, can easily be
accomplished in polynomial time using a nondeterministic algorithm. Step 2,
querying the satisfiability oracle, is done in one step. It follows that minimum
equivalent expression is in Σ
p
2
= NP
NP
.
11.10 Exercises
11.1. Show that the language in Example 11.1 is in DTIME(n). (Hint: Use a
2-tape Turing machine).
11.2. Show that the language L = ¦ww [ w ∈ ¦a, b¦
+
¦ is in LOGSPACE by
constructing a log space bounded oﬀ-line Turing machine that recognizes
L. Here ¦a, b¦
+
denotes all nonempty strings over the alphabet ¦a, b¦.
Exercises 329
11.3. Consider the following decision problem of sorting: Given a sequence of
n distinct positive integers between 1 and n, are they sorted in increasing
order? Show that this problem is in
(a) DTIME(nlog n).
(b) LOGSPACE.
11.4. Give an algorithm to solve the problem k-clique deﬁned in Exam-
ple 11.5. Use the O-notation to express the time complexity of your
algorithm.
11.5. Show that the problem k-clique deﬁned in Example 11.5 is in
LOGSPACE.
11.6. Show that the problem gap is in LOGSPACE if the graph is undirected.
11.7. Consider the following decision problem of the selection problem. Given
an array A[1..n] of integers, an integer x and an integer k, 1 ≤ k ≤ n, is
the kth smallest element in A equal to x? Show that this problem is in
LOGSPACE.
11.8. Let A be an n n matrix. Show that computing A
2
is in LOGSPACE.
How about computing A
k
for an arbitrary k ≥ 3, where k is part of the
input?
11.9. Show that the problem 2-SAT described on page 282 is in NLOGSPACE.
Conclude that it is in P.
11.10. Show that all ﬁnite sets are in LOGSPACE.
11.11. Show that the family of sets accepted by ﬁnite state automata is a proper
subset of LOGSPACE. (Hint: The language ¦a
n
b
n
[ n ≥ 1¦ is not a
ccepted by any ﬁnite state automaton, but it is in LOGSPACE.
11.12. Show that if T
1
and T
2
are two time-constructible functions, then so are
T
1
+T
2
, T
1
T
2
and 2
T
1
.
11.13. Prove Corollary 11.5.
11.14. Show that if NSPACE(n) ⊆ NP then NP = NSPACE. Conclude that
NSPACE(n) ,= NP.
11.15. Show that if LOGSPACE = NLOGSPACE, then for every space con-
structible function S(n) ≥ log n, DSPACE(S(n)) = NSPACE(S(n)).
11.16. Describe a log space reduction from the set L = ¦www [ w ∈ ¦a, b¦
+
¦ to
the set L

= ¦ww [ w ∈ ¦a, b¦
+
¦. That is, show that L ∝
log
L

.
11.17. Show that the relation ∝
poly
is transitive. That is, if Π ∝
poly
Π

and
Π

∝
poly
Π

, then Π ∝
poly
Π

.
11.18. Show that the relation ∝
log
is transitive. That is, if Π ∝
log
Π

and
Π

∝
log
Π

, then Π ∝
log
Π

.
330 Introduction to Computational Complexity
11.19. The problems 2-coloring and 2-SAT were deﬁned in Sec. 10.2. Show
that 2-coloring is log space reducible to 2-SAT. (Hint: Let G = (V, E).
Let the boolean variable x
v
correspond to vertex v for each vertex v ∈ V ,
and for each edge (u, v) ∈ E construct the two clauses (x
u
∨ x
v
) and
(x
u
∨ x
v
)).
11.20. A graph V = (G, E) is bipartite if and only if V can be partitioned into
two sets X and Y such that all edges in E are of the form (x, y) with
x ∈ X and y ∈ Y . Equivalently, G is bipartite if and only if it does not
contain odd-length cycles (see Sec. 3.3). Show that deciding whether
a graph is bipartite is log space reducible to the problem 2-coloring
described in Sec. 10.2.
11.21. Show that for any k ≥ 1, DTIME(n
k
) is not closed under polynomial
time reductions.
11.22. Show that, for any k ≥ 1, the class DSPACE(log
k
n) is closed under log
space reductions.
11.23. A set S is linear time reducible to a set T, denoted by S ∝
n
T, if there
exists a function f that can be computed in linear time (that is, f(x)
can be computed in c[x[ steps, for all input strings x, where c is some
constant > 0) such that
∀x x ∈ S if and only if f(x) ∈ T.
Show that if S ∝
n
T and T is in DTIME(n
k
), then S is in DTIME(n
k
).
That is, DTIME(n
k
) (k ≥ 1) is closed under linear time reducibility.
11.24. Suppose that k in Exercise 11.5 is not ﬁxed, that is, k is part of the
input. Will the problem still be in LOGSPACE? Explain.
11.25. Show that the class NLOGSPACE is closed under complementation.
Conclude that the complement of the problem gap is NLOGSPACE-
complete.
11.26. Show that the problem gap remains NLOGSPACE-complete even if the
graph is acyclic.
11.27. Show that the problem 2-SAT described in Sec. 10.2 is complete for the
class NLOGSPACE under log space reduction (see Exercise 11.9). (Hint:
Reduce the complement of the problem gap to it. Let G = (V, E) be a
directed acyclic graph. gap is NLOGSPACE-complete even if the graph
is acyclic (Exercise 11.26). By Exercise 11.25, the complement of the
problem gap is NLOGSPACE-complete. Associate with each vertex v
in V a boolean variable x
v
. Associate with each edge (u, v) ∈ E the
clause (x
u
∨ x
v
), and add the clauses (x
s
) for the start vertex and
(x
t
) for the goal vertex t. Prove that 2-SAT is satisﬁable if and only if
there is no path from s to t).
Exercises 331
11.28. Deﬁne the class
POLYLOGSPACE =
¸
k≥1
DSPACE(log
k
n).
Show that there is no set that is complete for the class POLY-
LOGSPACE. (Hint: The class DSPACE(log
k
n) is closed under log space
reduction).
11.29. Prove that PSPACE ⊆ P if and only if PSPACE ⊆ PSPACE(n).
(Hint: Use padding argument).
11.30. Does there exist a problem that is complete for the class DTIME(n)
under log space reduction? Prove your answer.
11.31. Use the fact that there is an NP-complete problem to show that there
is no problem that is complete for the class NTIME(n
k
) under log space
reduction for any k ≥ 1.
11.32. Let / be a class that is closed under complementation and let the set L
(that is not necessarily in /) be such that
∀L

∈ / L

∝ L.
Show that
∀L

∈ co-/ L

∝ L.
11.33. Show that for any class of languages /, if L is complete for the class /,
then L is complete for the class co-/.
11.34. Show that NLOGSPACE is strictly contained in PSPACE.
11.35. Show that DEXT ,= PSPACE. (Hint: Show that DEXT is not closed
under ∝
poly
).
11.36. Prove
(a) Theorem 11.15(1).
(b) Theorem 11.15(2).
(c) Theorem 11.15(3).
11.37. Prove
(a) Theorem 11.16(1).
(b) Theorem 11.16(2).
11.38. Prove
(a) Theorem 11.17(1).
(b) Theorem 11.17(2).
11.39. Prove Theorem 11.18.
11.40. Show that the problem path system accessibility∈ P.
332 Introduction to Computational Complexity
11.41. Show that polynomial time Turing reduction as deﬁned on page 325
implies polynomial time transformation as deﬁned in Sec. 11.7. Is the
converse true? Explain.
11.42. Consider the max-clique problem deﬁned as follows. Given a graph G =
(V, E) and a positive integer k, decide whether the maximal complete
subgraph of G is of size k. Show that max-clique is in ∆
p
2
.
11.43. Prove that Σ
p
1
= NP.
11.44. Show that if Σ
p
k
⊆ Π
p
k
, then Σ
p
k
= Π
p
k
.
11.45. Let PLOGSPACE be the class of sets accepted by a parallel model of
computation using logarithmic space. Show that PLOGSPACE = P.
11.11 Bibliographic notes
Part of the material in this chapter is based on Sudborough(1982). Other
references include Balcazar, Diaz and Gabarro (1988,1990), Bovet and
Crescenzi (1994), Garey and Johnson (1979), Hopcroft and Ullman (1979)
and Papadimitriou(1994). The book by Bovet and Crescenzi (1994) pro-
vides a good introduction to the ﬁeld of computational complexity. The
ﬁrst attempt to make a systematic approach to computational complexity
was made by Rabin(1960). The study of time and space complexity can be
said to begin with Hartmanis and Stearns(1965), Stearns, Hartmanis and
Lewis(1965) and Lewis, Stearns and Harmanis(1965). This work contains
most of the basic theorems of complexity classes and time and space hierar-
chy. Theorem 11.4 is due to Savitch (1970). Extensive research in this ﬁeld
emerged and enormous number of papers have been published since then.
For comments about NP-complete problems, see the bibliographic notes
of Chapter 10. PSPACE-complete problems were ﬁrst studied in Karp
(1972) including csg recognition and lba acceptance. quantified
boolean formulas was shown to be PSPACE-complete in Stockmeyer
and Meyer (1973) and Stockmeyer (1974). The linear bounded automata
problem, which predates the NP = P question, is the problem of decid-
ing whether nondeterministic LBA’s are equivalent to deterministic LBA’s,
that is whether NSPACE(n) = DSPACE(n).
NLOGSPACE-complete problems were studied by Savitch (1970), Sud-
borough (1975a,b), Springsteel (1976), Jones (1975), and Jones, Lien and
Lasser (1976). The NLOGSPACE-completeness of the graph accissibil-
ity problem (gap) was proven in Jones (1975).
Bibliographic notes 333
P-complete problems under log space reductions were considered by
Cook (1973), Cook and Sethi (1976), Jones (1975). Jones and Lasser (1976)
contains a collection of P-complete problems. The P-completeness of path
system accessibility with respect to log space reduction was proven in
Cook (1974). In Lander (1975), the circuit value problem was proven to
be P-complete with respect to log space reduction. The P-completeness of
the max-flow problem with respect to log space reduction is due to Gold-
schlager, Shaw, and Staples (1982). The problem ordered depth-first
search was presented in Reif (1985). linear programming is proved
to be P-complete under log space reduction in Dobkin, Lipton and Reiss
(1979). The deﬁnition of P-complete problems in terms of NC-reductions
is due to Cook (1985).
The polynomial hierarchy was ﬁrst studied in Stockmeyer (1976). See
also Wrathall (1976) for complete problems.
Very detailed bibliographic notes, as well as more recent topics in the
ﬁeld of computational complexity including the study of probabilistic algo-
rithms, parallel algorithms and interactive proof systems can be found in
recent books on computational complexity cited above. See the book by
Greenlaw, Hoover and Ruzzo (1995) for a thorough exposition of the theory
of P-completeness, including an extensive list of P-complete problems.
334
Chapter 12
Lower Bounds
12.1 Introduction
When we described algorithms in the previous chapters, we analyzed their
time complexities, mostly in the worst case. We have occasionally charac-
terized a particular algorithm as being “eﬃcient” in the sense that it has
the lowest possible time complexity. In Chapter 1, we have denoted by
an optimal algorithm an algorithm for which both the upper bound of the
algorithm and the lower bound of the problem are asymptotically equiva-
lent. For virtually all algorithms we have encountered, we have been able to
ﬁnd an upper bound on the amount of computation the algorithm requires.
But the problem of ﬁnding a lower bound of a particular problem is much
harder, and indeed there are numerous problems whose lower bound is un-
known. This is due to the fact that when considering the lower bound of a
problem, we have to establish a lower bound on all algorithms that solve
that problem. This is by no means an easy task compared with computing
the worst case running time of a given algorithm. It turns out, however,
that most of the known lower bounds are either trivial or derived using a
model of computation that is severely constrained, in the sense that it is
not capable of performing some elementary operations, e.g. multiplication.
12.2 Trivial Lower Bounds
In this section, we consider those lower bounds that can be deduced using
intuitive argument without resorting to any model of computation or doing
sophisticated mathematics. We will give two examples of establishing trivial
335
336 Lower Bounds
lower bounds.
Example 12.1 Consider the problem of ﬁnding the maximum in a list of n
numbers. Clearly, every element in the list must be inspected, assuming that the
list is unordered. This means that we must spend at least Ω(1) time for each
element. It follows that any algorithm to ﬁnd the maximum in an unordered list
must spend Ω(n) time.
Example 12.2 Consider the problem of matrix multiplication. Any algorithm
to multiply two n n matrices must compute exactly n
2
values. Since at least
Ω(1) time must be spent in each evaluation, the time complexity of any algorithm
for multiplying two n n matrices is Ω(n
2
).
12.3 The Decision Tree Model
There are certain problems where it is realistic to consider the branching
instruction as the basic operation (see Deﬁnition 1.6). Thus, in this case the
number of comparisons becomes the primary measure of complexity. In the
case of sorting, for example, the output is identical to the input except for
order. Therefore, it becomes reasonable to consider a model of computation
in which all steps are two-way branches, based on a comparison between
two quantities. The usual representation of an algorithm consisting solely
of branches is a binary tree called a decision tree.
Let Π be a problem for which a lower bound is sought, and let the size
of an instance of Π be represented by a positive integer n. Then, for each
pair of algorithm and value of n, there is a corresponding decision tree that
“solves” instances of the problem of size n. As an example, Fig. 1.2 shows
two decision trees corresponding to Algorithm binarysearch on instances
of size 10 and 14, respectively.
12.3.1 The search problem
In this section, we derive a lower bound on the search problem: Given an
array A[1..n] of n elements, determine whether a given element x is in the
array. In Chapter 1, we have presented Algorithm linearsearch to solve
this problem. We have also presented Algorithm binarysearch for the
case when the list is sorted.
In the case of searching, each node of the decision tree corresponds to a
The Decision Tree Model 337
decision. The test represented by the root is made ﬁrst and control passes
to one of its children depending on the outcome. If the element x being
searched for is less than the element corresponding to an internal node,
control passes to its left child. If it is greater, then control passes to its
right child. The search ceases if x is equal to the element corresponding to
a node, or if the node is a leaf.
Consider ﬁrst the case when the list is not sorted. As shown in Exam-
ple 12.1 for ﬁnding a lower bound for the problem of ﬁnding the maximum
of a given list of elements, it is also true here that n − 1 comparisons are
both necessary and suﬃcient in the worst case. It follows that the problem
of searching an arbitrary list requires at least Ω(n) time in the worst case,
and hence Algorithm linearsearch is optimal.
As regards the case when the list is sorted, we argue as follows. Let A
be an algorithm for searching a sorted list with n elements, and consider
the decision tree T associated with A and n. Let the number of nodes in
T be m. We observe that m ≥ n. We also observe that the number of
comparisons performed in the worst case must correspond to the longest
path from the root of T to a leaf plus one. This is exactly the height of T
plus one. By Observation 3.3, the height of T is at least log n|. It follows
that the number of comparisons performed in the worst case is log n| +1.
This implies the following theorem:
Theorem 12.1 Any algorithm that searches a sorted sequence of n ele-
ments must perform at least log n| + 1 comparisons in the worst case.
By the above theorem and Theorem 1.1, we conclude that Algorithm
binarysearch is optimal.
12.3.2 The sorting problem
In this section, we derive a lower bound on the problem of sorting by com-
parisons. All sorting problems that are not comparison-based, e.g. radix
sort and bucket sort are excluded. In the case of sorting, each internal ver-
tex of the tree represents a decision, and each leaf corresponds to an output.
The test represented by the root is made ﬁrst and control passes to one of
its children depending on the outcome. The desired output is available at
the leaf reached. With each pair of sorting algorithm and value of n repre-
senting the number of elements to be sorted, we associate a decision tree.
338 Lower Bounds
Thus, for a ﬁxed value of n, the decision tree corresponding to mergesort,
for example, is diﬀerent from that of heapsort or insertionsort. If the
elements to be sorted are a
1
, a
2
, . . . , a
n
, then the output is a permutation of
these elements. It follows that any decision tree for the sorting problem
must have at least n! leaves. Figure 12.1 shows an example of a decision
tree for an algorithm that sorts three elements.
1:2
2:3
1:3
1:3
2:3
o
1
o
2
o
3
< < o
2
o
1
o
3
< <
o
1
o
3
o
2
< < o
3
o
1
o
2
< <
o
3
o
2
o
1
< < o
2
o
3
o
1
< <
Fig. 12.1 A decision tree for sorting three elements.
Clearly, the time complexity in the worst case is the length of a longest
path from the root to a leaf, which is the height of the decision tree.
Lemma 12.1 Let T be a binary tree with at least n! leaves. Then, the
height of T is at least nlog n −1.5n = Ω(nlog n).
Proof. Let l be the number of leaves in T, and let h be its height. By
Observation 3.1, the number of vertices at level h, which are leaves, is at
most 2
h
. Since l ≥ n!, we have
n! ≤ l ≤ 2
h
.
Consequently, h ≥ log n!. By Eq. 2.18 (page 81),
h ≥ log n! =
n
¸
j=1
log j ≥ nlog n −nlog e + log e ≥ nlog n −1.5n. ⁄
Lemma 12.1 implies the following important theorem:
The Algebraic Decision Tree Model 339
Theorem 12.2 Any comparison-based algorithm that sorts n elements
must perform Ω(nlog n) element comparisons in the worst case.
In Chapter 6, we have shown that if n is a power of 2, then Algorithm
mergesort performs nlog n −n +1 comparisons in the worst case, which
is very close to the lower bound in Lemma 12.1. In other words, the lower
bound we have obtained is almost achievable by Algorithm mergesort.
12.4 The Algebraic Decision Tree Model
The decision tree model as described in Sec. 12.3 is severely restricted, as it
only allows a comparison between two elements as the primary operation.
If the decision at each internal vertex is a comparison of a polynomial of
the input variables with the number 0, then the resulting decision tree is
called an algebraic decision tree. This model of computation is far more
powerful than the decision tree model, and in fact attains the power of the
RAM model of computation. When establishing lower bounds for decision
problems using this model, we usually ignore all arithmetic operations and
conﬁne our attention to the number of branching instructions. Thus, this
model is similar to the decision tree model in the sense that it is best suited
for combinatorial algorithms that deal with rearrangements of elements.
We deﬁne this model of computation more formally as follows.
An algebraic decision tree on a set of n variables x
1
, x
2
, . . . , x
n
is a
binary tree with the property that each vertex is labeled with a statement
in the following way. Associated with every internal vertex is a statement
that is essentially a test of the form: If f(x
1
, x
2
, . . . , x
n
) : 0 then branch
to the left child, else branch to the right child. Here “:” stands for any
comparison relation from the set ¦=, <, ≤¦. On the other hand, one of the
answers yes or no is associated with each leaf vertex.
An algebraic decision tree is of order d, for some integer d ≥ 1, if all
polynomials associated with the internal nodes of the tree have degree at
most d. If d = 1, i.e., if all polynomials at the internal vertices of an
algebraic decision tree are linear, then it is called a linear algebraic decision
tree (or simply linear decision tree). Let Π be a decision problem whose
input is a set of n real numbers x
1
, x
2
, . . . , x
n
. Then, associated with Π is a
subset W of the n-dimensional space E
n
such that a point (x
1
, x
2
, . . . , x
n
)
is in W if and only if the answer to the problem Π when presented with the
input x
1
, x
2
, . . . , x
n
is yes. We say that an algebraic decision tree T decides
340 Lower Bounds
the membership in W if whenever the computation starts at the root of T
with some point p = (x
1
, x
2
, . . . , x
n
), control eventually reaches a yes leaf
if and only if (x
1
, x
2
, . . . , x
n
) ∈ W.
As in the decision tree model, to derive a lower bound on the worst
case time complexity of a problem Π, it suﬃces to derive a lower bound on
the height of the algebraic decision tree that solves Π. Now, let W be the
subset of the n-dimensional space E
n
that is associated with the problem
Π. Suppose that in some way the number #W of the connected components
of the set W is known. We want to derive a lower bound on the height of
the algebraic decision tree for Π in terms of #W. We now establish this
relation for the case of linear decision trees.
Let T be a linear decision tree. Then every path from the root to a leaf
in T corresponds to a sequence of conditions having one of the following
forms:
f(x
1
, x
2
, . . . , x
n
) = 0, g(x
1
, x
2
, . . . , x
n
) < 0, and h(x
1
, x
2
, . . . , x
n
) ≤ 0.
Note that each of these functions is linear since we have assumed that T
is a linear decision tree. Thus, when the root of T is presented with a
point (x
1
, x
2
, . . . , x
n
), control eventually reaches a leaf l if and only if all
conditions on the path from the root to l are satisﬁed. By the linearity of
these conditions, the leaf l corresponds to an intersection of hyperplanes,
open halfspaces and closed halfspaces, i.e., it corresponds to a convex set.
Since this set is convex, it is necessarily connected, i.e., it consists of exactly
one component. Thus, each yes-leaf corresponds to exactly one connected
component. It follows that the number of leaves of T is at least #W. By
an argument similar to that in the proof of Lemma 12.1, the height of the
tree is at least log(#W)|. This implies the following theorem:
Theorem 12.3 Let W be a subset of E
n
, and let T be a linear decision
tree of n variables that accepts the set W. Then, the height of T is at
least log(#W)|.
The linear decision tree model is certainly very restricted. Therefore, it
is desirable to extend it to the more general algebraic decision tree model.
It turns out, however, that in this model, the above argument no longer
applies; a yes-leaf may have associated with it many connected components.
In this case, more complex mathematical analysis leads to the following
theorem:
The Algebraic Decision Tree Model 341
Theorem 12.4 Let W be a subset of E
n
, and let d be a ﬁxed positive
integer. Then, the height of any order d algebraic decision tree T that
accepts W is Ω(log #W −n).
One of the most important combinatorial problems is that of sorting a
set of n real numbers using only the operation of comparisons. We have
shown that under the decision tree model, this problem requires Ω(nlog n)
comparisons in the worst case. It can be shown that this bound is still valid
under many computational models, and in particular the algebraic decision
tree model of computation. We state this fact as a theorem.
Theorem 12.5 In the algebraic decision tree model of computation, sort-
ing n real numbers requires Ω(nlog n) element comparisons in the worst
case.
12.4.1 The element uniqueness problem
The problem element uniqueness is stated as follows. Given a set of n
real numbers, decide whether two of them are equal. We will now obtain a
lower bound on the time complexity of this problem using the algebraic de-
cision tree model of computation. A set of n real numbers ¦x
1
, x
2
, . . . , x
n
¦
can be viewed as a point (x
1
, x
2
, . . . , x
n
) in the n-dimensional space E
n
.
Let W ⊆ E
n
be the membership set of the problem element unique-
ness on ¦x
1
, x
2
, . . . , x
n
¦. In other words, W consists of the set of points
(x
1
, x
2
, . . . , x
n
) with the property that no two coordinates of which are
equal. It is not hard to see that W contains n! disjoint connected compo-
nents. Speciﬁcally, each permutation π of ¦1, 2, . . . , n¦ corresponds to the
set of points in E
n
W
π
= ¦(x
1
, x
2
, . . . , x
n
) [ x
π(1)
< x
π(2)
< . . . < x
π(n)
¦.
Clearly,
W = W
1
∪ W
2
∪ . . . ∪ W
n!
.
Moreover, these subsets are connected and disjoint. Thus, #W = n!, and
as a result, the following theorem follows from Theorem 12.4.
Theorem 12.6 In the algebraic decision tree model of computation,
any algorithm that solves the element uniqueness problem requires
342 Lower Bounds
Ω(nlog n) element comparisons in the worst case.
12.5 Linear Time Reductions
For the problem element uniqueness, we were able to obtain a lower
bound using the algebraic decision tree model of computation directly by
investigating the problem and applying Theorem 12.4. Another approach
for establishing lower bounds is by the use of reductions. Let A be a problem
whose lower bound is known to be Ω(f(n)), where n = o(f(n)), e.g. f(n) =
nlog n. Let B be a problem for which we wish to establish a lower bound
of Ω(f(n)). We establish this lower bound for problem B as follows.
(1) Convert the input to A into a suitable input to problem B.
(2) Solve problem B.
(3) Convert the output into a correct solution to problem A.
In order to achieve a linear time reduction, Steps 1 and 3 above must be
performed in time O(n). In this case, we say that the problem A has been
reduced to the problem B in linear time, and we denote this by writing
A ∝
n
B.
Now we give examples of establishing an Ω(nlog n) lower bound for
three problems using the linear time reduction technique.
12.5.1 The convex hull problem
Let ¦x
1
, x
2
, . . . , x
n
¦ be a set of positive real numbers. We show that we
can use any algorithm for the convex hull problem to sort these num-
bers using only O(n) time for converting the input and output. Since the
sorting problem is Ω(nlog n), it follows that the convex hull problem
is Ω(nlog n) as well; otherwise we will be able to sort in o(nlog n) time,
contradicting Theorem 12.5.
With each real number x
j
, we associate a point (x
j
, x
2
j
) in the two
dimensional plane. Thus, all the n constructed points lie on the parabola
y = x
2
(see Fig. 12.2).
If we use any algorithm for the convex hull problem to solve the
constructed instance, the output will be a list of the constructed points
sorted by their x-coordinates. To obtain the sorted numbers, we traverse
Linear Time Reductions 343
Fig. 12.2 Reducing sorting to the convex hull problem.
the list and read oﬀ the ﬁrst coordinate of each point. The result is the
original set of numbers in sorted order. Thus we have shown that
sorting ∝
n
convex hull,
which proves the following theorem:
Theorem 12.7 In the algebraic decision tree model of computation, any
algorithm that solves the convex hull problem requires Ω(nlog n) oper-
ations in the worst case.
12.5.2 The closest pair problem
Given a set S of n points in the plane, the closest pair problem calls for
identifying a pair of points in S with minimum separation (see Sec. 6.9).
We show here that this problem requires Ω(nlog n) operations in the worst
case by reducing the problem element uniqueness to it.
Let ¦x
1
, x
2
, . . . , x
n
¦ be a set of positive real numbers. We show that
we can use an algorithm for the closest pair problem to decide whether
there are two numbers that are equal. Corresponding to each number x
j
,
we construct a point p
j
= (x
j
, 0). Thus the constructed set of points are
all on the line y = 0. Let A be any algorithm that solves the closest
pair problem. Let (x
i
, 0) and (x
j
, 0) be the output of algorithm A when
presented with the set of constructed points. Clearly, there are two equal
numbers in the original instance of the problem element uniqueness if
344 Lower Bounds
and only if the distance between x
i
and x
j
is equal to zero. Thus we have
shown that
element uniqueness ∝
n
closest pair.
This proves the following theorem:
Theorem 12.8 In the algebraic decision tree model of computation, any
algorithm that solves the closest pair problem requires Ω(nlog n) oper-
ations in the worst case.
12.5.3 The Euclidean minimum spanning tree problem
Let S be a set of n points in the plane. The euclidean minimum spanning
tree problem (emst) is to construct a tree of minimum total length whose
vertices are the given points in S. We show that this problem requires
Ω(nlog n) operations in the worst case by reducing the sorting problem
to it.
Let ¦x
1
, x
2
, . . . , x
n
¦ be a set of positive real numbers to be sorted. Cor-
responding to each number x
j
, we construct a point p
j
= (x
j
, 0). Thus
the constructed set of points are all on the line y = 0. Let A be any algo-
rithm that solves the euclidean minimum spanning tree problem. If we
feed algorithm A with the constructed set of points, the resulting minimum
spanning tree will consist of n −1 line segments l
1
, l
2
, . . . , l
n−1
on the line
y = 0, with the property that for each j, 1 ≤ j ≤ n−2, the right endpoint of
l
j
is the left endpoint of l
j+1
. We can obtain the numbers ¦x
1
, x
2
, . . . , x
n
¦
in sorted order by traversing the tree starting from the leftmost point and
reading oﬀ the ﬁrst component of each point. Thus we have shown that
sorting ∝
n
euclidean minimum spanning tree.
This proves the following theorem:
Theorem 12.9 In the algebraic decision tree model of computation, any
algorithm that solves the euclidean minimum spanning tree problem
requires Ω(nlog n) operations in the worst case.
Exercises 345
12.6 Exercises
12.1. Give trivial lower bounds for the following problems:
(a) Finding the inverse of an n n matix.
(b) Finding the median of n elements.
(c) Deciding whether a given array A[1..n] of n elements is sorted.
12.2. Draw the decision tree for Algorithm linearsearch on four elements.
12.3. Draw the decision tree for Algorithm insertionsort on three elements.
12.4. Draw the decision tree for Algorithm mergesort on three elements.
12.5. Let A and B be two unordered lists of n elements each. Consider the
problem of deciding whether the elements in A are identical to those in
B, i.e., the elements in A are a permutation of the elements in B. Use
the Ω-notation to express the number of comparisons required to solve
this problem.
12.6. What is the minimum number of comparisons needed to test whether an
array A[1..n] is a heap? Explain.
12.7. Let S be a list of n unsorted elements. Show that constructing a binary
search tree from the elements in S requires Ω(nlog n) in the decision tree
model (see Sec. 3.6.2 for the deﬁnition of a binary search tree).
12.8. Let S = ¦x
1
, x
2
, . . . , x
n
¦ be a set of n positive integers. We want to
ﬁnd an element x that is in the upper half when S is sorted, or in other
words an element that is greater than the median. What is the minimum
number of element comparisons required to solve this problem?
12.9. Let A[1..n] be an array of n integers in the range [1..m], where m > n.
We want to ﬁnd an integer x in the range [1..m] that is not in A. What
is the minimum number of comparisons required to solve this problem?
12.10. Let A
1
and A
2
be two sorted arrays of n elements each. Show that
at least 2n − 1 comparisons are required to merge A
1
and A
2
in the
worst case.
12.11. Referring to Exercise 12.10, construct two sorted arrays of n elements
each whose merging into one sorted array requires exactly 2n − 1 com-
parisons.
12.12. Give an algorithm to ﬁnd both the largest and second largest of an
unordered list of n elements. Your algorithm should perform the least
number of element comparisons.
12.13. Show that any algorithm for ﬁnding both the largest and second largest
of an unordered list of n elements, where n is a power of 2, must perform
at least n −2 + log n comparisons.
346 Lower Bounds
12.14. Consider the set disjointness problem: Given two sets of n real num-
bers each, determine whether they are disjoint. Show that any algorithm
to solve this problem requires Ω(nlog n) operations in the worst case.
12.15. Let A and B be two sets of points in the plane each containing n elements.
Show that the problem of ﬁnding two closest points, one in A and the
other in B requires Ω(nlog n) operations in the worst case.
12.16. Consider the triangulation problem: Given n points in the plane, join
them by nonintersecting straight line segments so that every region in-
ternal to their convex hull is a triangle. Prove that this problem requires
Ω(nlog n) operations in the worst case. (Hint: Reduce the sorting prob-
lem to the special case of the triangulation problem when exactly n−1
points are collinear and one point is not on the same line).
12.17. Consider the nearest point problem: Given a set S of n points in the
plane and a query point p, ﬁnd a point in S that is closest to p. Show
that any algorithm to solve this problem requires Ω(log n) operations in
the worst case. (Hint: Reduce binary search to the special case where
all points lie on the same line).
12.18. The all nearest points problem is deﬁned as follows. Given n points
in the plane, ﬁnd a nearest neighbor of each. Show that this prob-
lem requires Ω(nlog n) operations in the worst case. (Hint: Reduce the
closest pair problem to it).
12.19. Let S be a set of n points in the plane. The diameter of S, denoted by
Diam(S), is the maximum distance realized by two points in S. Show
that ﬁnding Diam(S) requires Ω(nlog n) operations in the worst case.
12.20. Consider the problem of partitioning a planar point set S into two sub-
sets S
1
and S
2
such that the maximum of Diam(S
1
) and Diam(S
2
) is
minimum. Show that this problem requires Ω(nlog n) operations in the
worst case. (Hint: Reduce the problem of ﬁnding the diameter of a point
set S to this problem; see Exercise 12.19).
12.7 Bibliographic notes
For a detailed account of lower bounds for sorting, merging and selection
see Knuth (1973). This book provides an in-depth analysis. A sorting algo-
rithm that requires the fewest known number of comparisons was originally
presented in Ford and Johnson (1959). A merging algorithm with the min-
imum number of comparisons was presented by Hwang and Lin (1972). A
lower bound on selection can be found in Hyaﬁl (1976). Other relevant pa-
pers containing lower bound results include Fussenegger and Gabow (1976),
Reingold (1971), Reingold (1972) and Friedman (1972). Theorem 12.3 is
Bibliographic notes 347
due to Dobkin and Lipton (1979). Theorem 12.4 is due to Bin-Or (1983).
Establishing lower bounds for many geometric problems can be found in
Preparata and Shamos (1985).
348
PART 5
Coping with Hardness
349
350
351
In the previous part of the book, we have seen that many practical prob-
lems have no eﬃcient algorithms, and the known ones for these problems
require an amount of time measured in years or centuries even for instances
of moderate size.
There are three useful methodologies that could be used to cope with
this diﬃculty. The ﬁrst methodology is suitable for those problems that
exhibit good average time complexity, but for which the worst case poly-
nomial time solution is elusive. This methodology is based on a methodic
examination of the implicit state space induced by the problem instance
under study. In the process of exploring the state space of the instance,
some pruning takes place.
The second methodology in this part is based on the probabilistic no-
tion of accuracy. At the heart of these solutions is a simple decision maker
or test that can accurately perform one task (either passing or failing the
alternative) and not say much about the complementary option. An iter-
ation through this test will enable the construction of the solution or the
increase in the conﬁdence level in the solution to the desired degree.
The ﬁnal methodology is useful for incremental solutions where one
is willing to compromise on the quality of solution in return for faster
(polynomial time) solutions. Only some classes of hard problems admit such
polynomial time approximations. Still fewer of those provide a specturum of
polynomial time solutions where the degree of the polynomial is a function
of accuracy.
In Chapter13, we study two solution space search techniques that work
for some problems, especially those in which the solution space is large.
These techniques are backtracking and branch-and-bound. In these tech-
niques, a solution to the problem can be obtained by exhaustively search-
ing through a large but ﬁnite number of possibilities. It turns out that
for many hard problems, backtracking and branch-and-bound are the only
known techniques to solve these problems. After all, for some problems like
the traveling salesman problem, even the problem of ﬁnding an approxi-
mate solution is NP-hard. In this chapter, a well-known branch-and-bound
algorithm for the traveling salesman problem is presented. Other examples
that are solved using the backtracking technique in this chapter include
3-coloring and the 8-queens problems.
Randomized algorithms are the subject of Chapter 14. In this chapter,
we ﬁrst show that randomization improves the performance of algorithm
quicksort signiﬁcantly, and results in a randomized selection algorithm
352
that is considerably simpler and (almost always) much faster than Algo-
rithm select discussed in Chapter 6. Next, we present two randomized
algorithms for pattern matching and sampling. Finally, we apply random-
ization to a hard problem in number theory: primality testing. We will
describe an eﬃcient algorithm for this problem that almost all the time
decides correctly whether a given positive integer is prime or not.
Chapter 15 discusses another avenue for dealing with hard problems:
Instead of obtaining an optimal solution, we may be content with an ap-
proximate solution. In this chapter, we study some approximation algo-
rithms for some NP-hard problems including the bin packing problem, the
Euclidean traveling salesman problem, the knapsack problem and the ver-
tex cover problem. These problems share the common feature that the ratio
of the optimal solution to the approximate solution is bounded by a small
(and reasonable) constant. For the knapsack problem, we show a polyno-
mial approximation scheme; that is an algorithm that receives as input the
desired approximation ratio and delivers an output whose relative ratio to
the optimal solution is within the input ratio. This kind of polynomial
approximation scheme is not polynomial in the reciprocal of the desired
ratio. For this reason, we extend this scheme to the fully polynomial time
approximation scheme that is also polynomial in the reciprocal of the de-
sired ratio. As an example of this technique, we present an approximation
algorithm for the subset sum problem.
Chapter 13
Backtracking
13.1 Introduction
In many real world problems, as in most of the NP-hard problems, a solution
can be obtained by exhaustively searching through a large but ﬁnite number
of possibilities. Moreover, for virtually all these problems, there does not
exist an algorithm that uses a method other than exhaustive search. Hence,
the need arose for developing systematic techniques of searching, with the
hope of cutting down the search space to possibly a much smaller space. In
this chapter, we present a general technique for organizing the search known
as backtracking. This algorithm design technique can be described as an
organized exhaustive search which often avoids searching all possibilities.
It is generally suitable for solving problems where a potentially large but
ﬁnite number of solutions have to be inspected.
13.2 The 3-Coloring Problem
Consider the problem 3-coloring: Given an undirected graph G = (V, E),
it is required to color each vertex in V with one of three colors, say 1, 2,
and 3, such that no two adjacent vertices have the same color. We call such
a coloring legal; otherwise, if two adjacent vertices have the same color, it
is illegal. A coloring can be represented by an n-tuple (c
1
, c
2
, . . . , c
n
) such
that c
i
∈ ¦1, 2, 3¦, 1 ≤ i ≤ n. For example, (1, 2, 2, 3, 1) denotes a coloring
of a graph with ﬁve vertices. There are 3
n
possible colorings (legal and
illegal) to color a graph with n vertices. The set of all possible colorings
can be represented by a complete ternary tree called the search tree. In
353
354 Backtracking
this tree, each path from the root to a leaf node represents one coloring
assignment. Figure 13.1 shows such a tree for the case of a graph with 3
vertices.
Fig. 13.1 The search tree for all possible 3-colorings for a graph with 3 vertices.
Let us call an incomplete coloring of a graph partial if no two adjacent
colored vertices have the same color. Backtracking works by generating the
underlying tree one node at a time. If the path from the root to the current
node corresponds to a legal coloring, the process is terminated (unless more
than one coloring is desired). If the length of this path is less than n and
the corresponding coloring is partial, then one child of the current node is
generated and is marked as the current node. If, on the other hand, the
corresponding path is not partial, then the current node is marked as a
dead node and a new node corresponding to another color is generated.
If, however, all three colors have been tried with no success, the search
backtracks to the parent node whose color is changed, and so on.
Example 13.1 Consider the graph shown in Fig. 13.2(a), where we are inter-
ested in coloring its vertices using the colors ¦1, 2, 3¦. Figure 13.2(b) shows part
of the search tree generated during the process of searching for a legal coloring.
First, after generating the third node, it is discovered that the coloring (1, 1) is
not partial, and hence that node is marked as a dead node by marking it with
in the ﬁgure. Next, b is assigned the color 2, and it is seen that the coloring (1,
2) is partial. Hence, a new child node corresponding to vertex c is generated with
an initial color assignment of 1. Repeating the above procedure of ignoring dead
nodes and expanding those corresponding to partial colorings, we ﬁnally arrive
at the legal coloring (1, 2, 2, 1, 3).
The 3-Coloring Problem 355
It is interesting to note that we have arrived at the solution after generating
only 10 nodes out of the 364 nodes comprising the search tree.
o=1
/=2
c=2
d=1
c=2
c=3
c=1
(b)
/=1
c=1
o
/ c
c d
(a)
Fig. 13.2 An example of using backtracking to solve the problem 3-coloring.
There are two important observations to be noted in Example 13.1,
which generalize to all backtracking algorithms. First, the nodes are gen-
erated in a depth-ﬁrst-search manner. Second, there is no need to store
the whole search tree; we only need to store the path from the root to the
current active node. In fact, no physical nodes are generated at all; the
whole tree is implicit. In our example above, we only need to keep track of
the color assignment.
The algorithm
Now we proceed to give two algorithms that use backtracking to solve the
3-coloring problem, one is recursive and the other is iterative. In both
algorithms, we assume for simplicity that the set of vertices is ¦1, 2, . . . , n¦.
The recursive algorithm is shown as Algorithm 3-colorrec.
Initially, no vertex is colored and this is indicated by setting all colors
to 0 in Step 1. The call graphcolor(1) causes the ﬁrst vertex to be colored
with 1. Clearly, (1) is a partial coloring, and hence the procedure is then
recursively called with k = 2. The assignment statement causes the second
vertex to be colored with 1 as well. The resulting coloring is (1, 1). If
356 Backtracking
Algorithm 13.1 3-colorrec
Input: An undirected graph G = (V, E).
Output: A 3-coloring c[1..n] of the vertices of G, where each c[j] is 1,2 or 3.
1. for k←1 to n
2. c[k] ←0
3. end for
4. ﬂag ←false
5. graphcolor(1)
6. if ﬂag then output c
7. else output “no solution”
Procedure graphcolor(k)
1. for color = 1 to 3
2. c[k] ←color
3. if c is a legal coloring then set ﬂag ←true and exit
4. else if c is partial then graphcolor(k + 1)
5. end for
vertices 1 and 2 are not connected by an edge, then this coloring is partial.
Otherwise, the coloring is not partial, and hence the second vertex will be
colored with 2 and the resulting coloring is (1, 2). After the second vertex
has been colored, i.e., if the current coloring is partial, the procedure is
again invoked with k = 3, and so on. Suppose that the procedure fails
to color vertex j for some vertex j ≥ 3. This happens if the for loop is
executed three times without ﬁnding a legal or partial coloring. In this case,
the previous recursive call is activated and another color for vertex j − 1
is tried. If again none of the three colors result in a partial coloring, the
one before the last recursive call is activated. This is where backtracking
takes place. The process of advancing and backtracking is continued until
the graph is either colored or all possibilities have been exhausted without
ﬁnding a legal coloring. Checking whether a coloring is partial can be done
incrementally: If the coloring vector c contains m nonzero numbers and
c[m] does not result in a conﬂict with any other color then it is partial;
otherwise it is not partial. Checking whether a coloring is legal amounts to
checking whether the coloring vector consists of noncontradictory n colors.
The iterative backtracking algorithm is given as Algorithm 3-
coloriter. The main part of this algorithm consists of two nested
while loops. The inner while loop implements advances (generating new
nodes), whereas the outer while loop implements backtracking (to previ-
The 8-Queens Problem 357
ously generated nodes). The working of this algorithm is similar to that of
the recursive version.
Algorithm 13.2 3-coloriter
Input: An undirected graph G = (V, E).
Output: A 3-coloring c[1..n] of the vertices of G, where each c[j] is 1,2 or 3.
1. for k←1 to n
2. c[k] ←0
3. end for
4. ﬂag ←false
5. k←1
6. while k ≥ 1
7. while c[k] ≤ 2
8. c[k] ←c[k] + 1
9. if c is a legal coloring then set ﬂag ←true
and exit from the two while loops.
10. else if c is partial then k←k + 1 ¦advance ¦
11. end while
12. c[k] ←0
13. k←k −1 ¦backtrack¦
14. end while
15. if ﬂag then output c
16. else output “no solution”
As to the time complexity of these two algorithms, we note that O(3
n
)
nodes are generated in the worst case. For each generated node, O(n)
work is required to check if the current coloring is legal, partial, or neither.
Hence, the overall running time is O(n3
n
) in the worst case.
13.3 The 8-Queens Problem
The classical 8-queens can be stated as follows. How can we arrange 8
queens on an 88 chessboard so that no two queens can attack each other?
Two queens can attack each other if they are in the same row, column or
diagonal. The n-queens problem is deﬁned similarly, where in this case we
have n queens and an n n chessboard for an arbitrary value of n ≥ 1.
To simplify the discussion, we will study the 4-queens problem, and the
generalization to any arbitrary n is straightforward.
Consider a chessboard of size 44. Since no two queens can be put in the
358 Backtracking
same row, each queen is in a diﬀerent row. Since there are four positions in
each row, there are 4
4
possible conﬁgurations. Each possible conﬁguration
can be described by a vector with four components x = (x
1
, x
2
, x
3
, x
4
).
For example, the vector (2, 3, 4, 1) corresponds to the conﬁguration shown
in Fig. 13.3(a). A component is zero (and hence not included explicitly
in the vector) if there is no queen placed in its corresponding row. For
example, the partial vector (3, 1) corresponds to the conﬁguration shown
in Fig. 13.3(b). In fact, since no two queens can be placed in the same
column, a legal placement corresponds to a permutation of the numbers 1,
2, 3 and 4. This reduces the search space from 4
4
to 4!. Modifying the
algorithm accordingly will be left as an exercise.
(a) (b)
Fig. 13.3 Two conﬁgurations of the 4-queens problem.
The algorithm
To solve the 8-queens problem using backtracking, the algorithm tries to
generate and search a complete 4-ary rooted tree in a depth-ﬁrst manner.
The root of the tree corresponds to the placement of no queens. The nodes
on the ﬁrst level correspond to the possible placements of the queen in the
ﬁrst row, those on the second level correspond to the possible placements
of the queen in the second row and so on. The backtracking algorithm to
solve this problem is given as Algorithm 4-queens. In the algorithm, we
used the term legal to mean a placement of 4 queens that do not attack
each other, and the term partial to mean a placement of less than 4 queens
that do not attack each other. Clearly, two queens placed at positions x
i
and x
j
are in the same column if and only if x
i
= x
j
. It is not hard to see
The 8-Queens Problem 359
that two queens are in the same diagonal if and only if
x
i
−x
j
= i −j or x
i
−x
j
= j −i.
Algorithm 13.3 4-queens
Input: none.
Output: A vector x[1..4] corresponding to the solution of the 4-queens prob-
lem.
1. for k←1 to 4
2. x[k] ←0 ¦no queens are placed on the chessboard ¦
3. end for
4. ﬂag ←false
5. k←1
6. while k ≥ 1
7. while x[k] ≤ 3
8. x[k] ←x[k] + 1
9. if x is a legal placement then set ﬂag ←true
and exit from the two while loops.
10. else if x is partial then k←k + 1 ¦advance¦
11. end while
12. x[k] ←0
13. k←k −1 ¦backtrack¦
14. end while
15. if ﬂag then output x
16. else output “no solution”
Example 13.2 Applying the algorithm produces the solution shown in
Fig. 13.4. In the ﬁgure, deadend nodes are marked with . First, x
1
is set
to 1 and x
2
is set to 1. This results in a deadend, as the two queens are in the
same column. The same result happens if x
2
is set to 2 since in this case the
two queens are on the same diagonal. Setting x
2
to 3 results in the partial vector
(1, 3) and the search advances to ﬁnd a value for x
3
. As shown in the ﬁgure, no
matter what value x
3
assumes, no partial vector results with x
1
= 1, x
2
= 3 and
x
3
> 0. Hence, the search backtracks to the second level and x
2
is reassigned a
new value, namely 4. As shown in the ﬁgure, this results in the partial vector
(1, 4, 2). Again, this vector cannot be extended and consequently, after generating
a few nodes, the search backs up to the ﬁrst level. Now, x
1
is incremented to 2
and, in the same manner, the partial vector (2, 4, 1) is found. As shown in the
ﬁgure, this vector is extended to the legal vector (2, 4, 1, 3), which corresponds to
a legal placement.
360 Backtracking
a = 2
1
a = 4
2
a = 1
3
a = 3
4
Fig. 13.4 An example of using backtracking to solve the 4-queens problem.
Now, consider a brute-force method to solve the general n-queens prob-
lem. As mentioned before, since no two queens can be placed in the
same column, the solution vector must be a permutation of the numbers
1, 2, . . . , n. Thus, the brute-force method can be improved to test n! con-
ﬁgurations instead of n
n
. However, the following argument shows that
backtracking drastically cuts the number of tests. Consider the (n − 2)!
vectors corresponding to those conﬁgurations in which the ﬁrst two queens
are placed in the ﬁrst column. The brute-force method blindly tests all
these vectors, whereas in backtracking these tests can be avoided using
O(1) tests. Although the backtracking method to solve the n-queens prob-
lem costs O(n
n
) in the worst case, it empirically far exceeds the O(n!)
brute-force method in eﬃciency, as its expected running time is generally
much faster. For example, the algorithm discovered the solution shown in
Fig. 13.4 after generating 27 nodes out of a total of 341 possible nodes.
13.4 The General Backtracking Method
In this section we describe the general backtracking algorithm as a system-
atic search method that can be applied to a class of search problems whose
solution consists of a vector (x
1
, x
2
, . . . , x
i
) satisfying some predeﬁned con-
straints. Here i is some integer between 0 and n, where n is a constant that
is dependent on the problem formulation. In the two algorithms we have
The General Backtracking Method 361
covered, 3-coloring and the 8-queens problems, i was ﬁxed. However,
in some problems i may vary from one solution to another as the following
example illustrates.
Example 13.3 Consider a variant of the partition problem deﬁned as fol-
lows. Given a set of n integers X = ¦x
1
, x
2
, . . . , x
n
¦ and an integer y, ﬁnd a
subset Y of X whose sum is equal to y. For instance if
X = ¦10, 20, 30, 40, 50, 60¦
and y = 60, then there are, for example,three solutions of diﬀerent lengths, namely
¦10, 20, 30¦, ¦20, 40¦, and ¦60¦.
It is not hard to devise a backtracking algorithm to solve this problem. Note
that this problem can be formulated in another way so that the solution is a
boolean vector of length n in the obvious way. Thus, the above three solutions
may be expressed by the boolean vectors
¦1, 1, 1, 0, 0, 0, ¦, ¦0, 1, 0, 1, 0, 0¦, and ¦0, 0, 0, 0, 0, 1¦.
In backtracking, each x
i
in the solution vector belongs to a ﬁnite linearly
ordered set X
i
. Thus, the backtracking algorithm considers the elements of
the cartesian product X
1
X
2
. . .X
n
in lexicographic order. Initially, the
algorithm starts with the empty vector. It then chooses the least element of
X
1
as x
1
. If (x
1
) is a partial solution, the algorithm proceeds by choosing
the least element of X
2
as x
2
. If (x
1
, x
2
) is a partial solution, then the least
element of X
3
is included; otherwise x
2
is set to the next element in X
2
.
In general, suppose that the algorithm has detected the partial solution
(x
1
, x
2
, . . . , x
j
). It then considers the vector v = (x
1
, x
2
, . . . , x
j
, x
j+1
). We
have the following cases:
(1) If v represents a ﬁnal solution to the problem, the algorithm records
it as a solution and either terminates in case only one solution is
desired or continues to ﬁnd other solutions.
(2) (The advance step). If v represents a partial solution, the algorithm
advances by choosing the least element in the set X
j+2
.
(3) If v is neither a ﬁnal nor a partial solution, we have two subcases:
(a) If there are still more elements to choose from in the set X
j+1
,
the algorithm sets x
j+1
to the next member of X
j+1
.
362 Backtracking
(b) (The backtrack step) If there are no more elements to choose
from in the set X
j+1
, the algorithm backtracks by setting x
j
to
the next member of X
j
. If again there are no more elements to
choose from in the set X
j
, the algorithm backtracks by setting
x
j−1
to the next member of X
j−1
, and so on.
Now, we describe the general backtracking algorithm formally using
two procedures: one recursive (backtrackrec) and the other iterative
(backtrackiter).
Algorithm 13.4 backtrackrec
Input: Explicit or implicit description of the sets X
1
, X
2
, . . . , X
n
.
Output: A solution vector v = (x
1
, x
2
, . . . , x
i
), 0 ≤ i ≤ n.
1. v←( )
2. ﬂag ←false
3. advance(1)
4. if ﬂag then output v
5. else output “no solution”
Procedure advance(k)
1. for each x ∈ X
k
2. x
k
← x; append x
k
to v
3. if v is a ﬁnal solution then set ﬂag ←true and exit
4. else if v is partial then advance(k + 1)
5. end for
These two algorithms very much resemble those backtracking algorithms
described in Secs. 13.2 and 13.3. In general, to search for a solution to a
problem using backtracking, one of these two prototype algorithms may be
utilized as a framework around which an algorithm specially tailored to the
problem at hand can be designed.
13.5 Branch and Bound
Branch and bound design technique is similar to backtracking in the sense
that it generates a search tree and looks for one or more solutions. However,
while backtracking searches for a solution or a set of solutions that satisfy
certain properties (including maximization or minimization), branch-and-
bound algorithms are typically concerned with only maximization or min-
Branch and Bound 363
Algorithm 13.5 backtrackiter
Input: Explicit or implicit description of the sets X
1
, X
2
, . . . , X
n
.
Output: A solution vector v = (x
1
, x
2
, . . . , x
i
), 0 ≤ i ≤ n.
1. v←( )
2. ﬂag ←false
3. k←1
4. while k ≥ 1
5. while X
k
is not exhausted
6. x
k
← next element in X
k
; append x
k
to v
7. if v is a ﬁnal solution then set ﬂag ←true
and exit from the two while loops.
8. else if v is partial then k←k + 1 ¦advance¦
9. end while
10. Reset X
k
so that the next element is the ﬁrst.
11. k←k −1 ¦backtrack¦
12. end while
13. if ﬂag then output v
14. else output “no solution”
imization of a given function. Moreover, in branch-and-bound algorithms,
a bound is calculated at each node x on the possible value of any solution
given by nodes that may later be generated in the subtree rooted at x. If
the bound calculated is worse than a previous bound, the subtree rooted
at x is blocked, i.e., none of its children are generated.
Henceforth, we will assume that the algorithm is to minimize a given
cost function; the case of maximization is similar. In order for branch
and bound to be applicable, the cost function must satisfy the following
property. For all partial solutions (x
1
, x
2
, . . . , x
k−1
) and their extensions
(x
1
, x
2
, . . . , x
k
), we must have
cost(x
1
, x
2
, . . . , x
k−1
) ≤ cost(x
1
, x
2
, . . . , x
k
).
Given this property, a partial solution (x
1
, x
2
, . . . , x
k
) can be discarded once
it is generated if its cost is greater than or equal to a previously computed
solution. Thus, if the algorithm ﬁnds a solution whose cost is c, and there is
a partial solution whose cost is at least c, no more extensions of this partial
solution are generated.
The traveling salesman problem will serve as a good example for
the branch and bound method. This problem is deﬁned as follows. Given
a set of cities and a cost function that is deﬁned on each pair of cities, ﬁnd
364 Backtracking
a tour of minimum cost. Here a tour is a closed path that visits each city
exactly once, i.e., a simple cycle. The cost function may be the distance,
travel time, air fare, etc. An instance of the traveling salesman is
given by its cost matrix whose entries are assumed to be nonnegative. The
matrix A in Fig. 13.5 is an example of such an instance. With each partial
solution (x
1
, x
2
, . . . , x
k
), we associate a lower bound y which is interpreted
as follows. The cost of any complete tour that visits the cities x
1
, x
2
, . . . , x
k
in this order must be at least y.
We observe that each complete tour must contain exactly one edge and
its associated cost from each row and each column of the cost matrix. We
also observe that if a constant r is subtracted from every entry in any row
or column of the cost matrix A, the cost of any tour under the new matrix
is exactly r less than the cost of the same tour under A. This motivates the
idea of reducing the cost matrix so that each row or column contains at least
one entry that is equal to 0. We will refer to such a matrix as the reduction
of the original matrix. In Fig. 13.5, matrix B is the reduction of A.
17 7 35 18
9 5 14 19
29 24 30 12
27 21 25 48
15 16 28 18
1 2 3 4 5
1
2
3
4
5
A
10 0 25 11
4 0 6 14
17 12 15 0
6 0 4 27
0 1 13 0
1 2 3 4 5
1
2
3
4
5
-7
-5
-12
-21
-15
-3
B
Fig. 13.5 An instance matrix of the traveling salesman and its reduction.
Matrix B in the ﬁgure results from subtracting the shown amounts
from each row and from column 4. The total amount subtracted is 63. It
is not hard to see that the cost of any tour is at least 63. In general, let
(r
1
, r
2
, . . . , r
n
) and (c
1
, c
2
, . . . , c
n
) be the amounts subtracted from rows 1
to n and columns 1 to n, respectively, in an n n cost matrix A. Then
y =
n
¸
i=1
r
i
+
n
¸
i=1
c
i
is a lower bound on the cost of any complete tour.
Now, we proceed to describe a branch-and-bound algorithm to solve the
Branch and Bound 365
traveling salesman problem through an example. Our example is to ﬁnd an
optimal tour for the instance given in Fig. 13.5. The search tree, which is
a binary tree, is depicted in Fig. 13.6.
(3,5)
(2,3)
(1,3)
(4,2)
0
(5,4)
solution
(2,1)
(3,5)
(1,3)
(4,2)
(2,3)
(2,1)
(5,4)
-12
10 0 25 11
4 0
6 14
5 0 3
6 0 4 27
0 1 13 0
1 2 3 4 5
1
2
3
4
5
bound = 75
D
B
10 0 25 11
4 0 6 14
17 12 15 0
6 0 4 27
0 1 13 0
1 2 3 4 5
1
2
3
4
5
bound = 63
6 0 4
0 1 0
4
5
10 0 25
4 0 6
1 2 3 4
1
2
bound = 63
C
6 0 4
0 1 0
4
5
10 0 25
0 2
1 2 3 4
1
2
bound = 67
-4
F
bound = 73
-10 0
1 2
1
6 0
0
4
5 0
15
4
E
bound = 73
2
-6
-1
0
1 2
2
0
0
4
5 0
4
J
0
1
2
bound = 67
0 5
4
I
1
2
bound = 67
0 5
4
L
bound = 67
5
4
K
bound = 67
5
4
M
0
1 2
2
bound = 67
6 0
1
4
5 0
2
4
G
-10
6 0 4
0 1 0
4
5
0 15
0 2
1 2 3 4
1
2
bound = 77
H
Fig. 13.6 Solution of the traveling salesman using branch and bound.
366 Backtracking
The root of the tree is represented by the reduction matrix B, and is
labeled with the lower bound computed above, namely 63. This node is
split into two nodes corresponding to the left and right subtrees. The right
subtree will contain all solutions that exclude the edge (3, 5) and thus the
entry D
3,5
is set to ∞. We will justify the choice of the edge (3, 5) later.
Since there are no zeros in row 3 of matrix D, it can be reduced further
by 12. This is accompanied by increasing the lower bound by 12 to become
75. The left subtree will contain all solutions that include the edge (3, 5)
and thus both the 3rd row and 5th column of matrix C are removed, since
we can never go from 3 to any other city nor arrive at 5 from any other
city. Furthermore, since all solutions in this subtree use the edge (3, 5), the
edge (5, 3) will not be used any more, and hence its corresponding entry
C
5,3
is set to ∞. As each row and column of this matrix contains a zero,
it cannot be reduced further and hence the lower bound of this node is the
same as its parent’s lower bound.
Now, as the lower bound of the node containing matrix C is less than
that of the node containing matrix D, the next split is performed on the
node containing matrix C. We use edge (2, 3) to split this node.
The right subtree will contain all solutions that exclude the edge (2, 3)
and thus the entry F
2,3
is set to ∞. Since there are no zeros in row 2 of
matrix F, it can be reduced further by 4. This increases the lower bound
from 63 to 67. The left subtree will contain all solutions that include the
edge (2, 3) and hence both the 2nd row and 3rd column of matrix E are
removed. Now, following the same procedure above, we would change E
3,2
to ∞. However, this entry does not exist in matrix E. If we follow the path
from the root to the node containing this matrix, we see that the two edges
(3, 5) and (2, 3), i.e., the subpath 2,3,5 must be in any tour in the subtree
whose root contains matrix E. This implies that the entry E
5,2
must be set
to ∞. In general, if the edge included is (u
i
, v
1
) and the path from the root
contains the two paths u
1
, u
2
, . . . , u
i
and v
1
, v
2
, . . . , v
j
, then M
v
j
,u
1
is set to
∞, where M is the matrix at the current node. To ﬁnish processing matrix
E, we subtract 10 from the ﬁrst row, which increases the lower bound from
63 to 73.
Following the above procedure, the matrices G, H, I, J, K, L, M are com-
puted in this order. The optimal tour can be traced from the root by fol-
lowing the lines shown in bold face, that is 1, 3, 5, 4, 2, 1. Its total cost is
7 + 12 + 18 + 21 + 9 = 67.
At the beginning of this example, we chose to split using the edge (3, 5)
Exercises 367
because it caused the greatest increase in the lower bound of the right
subtree. This heuristic is useful because it is faster to ﬁnd the solution
by following the left edges, which reduce the dimension as opposed to the
right edges which merely add a new ∞ and probably more zeros. However,
we did not use this heuristic when splitting at the node containing matrix
C. It is left as an exercise to ﬁnd the optimal solution with fewer node
splittings.
From the above example, it seems that the heap is an ideal data struc-
ture to use in order to expand the node with the least cost (or maximum
cost in case of maximization). Although branch-and-bound algorithms are
generally complicated and hard to program, they proved to be eﬃcient in
practice.
13.6 Exercises
13.1. Let k-coloring be a generalization of the 3-coloring problem pre-
sented in Sec. 13.2. How many nodes are generated by its corresponding
backtracking algorithm in the worst case?
13.2. Consider the algorithm for 3-coloring presented in Sec. 13.2. Give an
eﬃcient algorithm to test whether a vector corresponding to a 3-coloring
of a graph is legal.
13.3. Consider the algorithm for 3-coloring presented in Sec. 13.2. Explain
how to eﬃciently test whether the current vector is partial throughout
the execution of the algorithm.
13.4. Let Algorithm n-queens be a generalization of Algorithm 4-queens pre-
sented in Sec. 13.3 for the case of an nn chessboard. How many nodes
are generated by Algorithm n-queens in the worst case?
13.5. Show that two queens placed at positions x
i
and x
j
are in the same
diagonal if and only if
x
i
−x
j
= i −j or x
i
−x
j
= j −i.
13.6. Give a recursive algorithm for the 8-queens problem.
13.7. Does the n-queen problem have a solution for every value of n ≥ 4?
Prove your answer.
13.8. Modify Algorithm 4-queens so that it reduces the search space from 4
4
to 4! as described in Sec. 13.3.
13.9. Design a backtracking algorithm to generate all permutations of the num-
bers 1, 2, . . . , n.
368 Backtracking
13.10. Design a backtracking algorithm to generate all 2
n
subsets of the numbers
1, 2, . . . , n.
13.11. Write a backtracking algorithm to solve the knight tour problem: Given
an 88 chessboard, decide if it is possible for a knight placed at a certain
position of the board to visit every square of the board exactly once and
return to its start position.
13.12. Write a backtracking algorithm to solve the following variant of the par-
tition problem (see Example 13.3): Given n positive integers X =
¦x
1
, x
2
, . . . , x
n
¦ and a positive integer y, does there exist a subset Y ⊆ X
whose elements sum up to y?
13.13. Give a backtracking algorithm to solve the Hamiltonian cycle prob-
lem: Given an undirected graph G = (V, E), determine whether it con-
tains a simple cycle that visits each vertex exactly once.
13.14. Consider the knapsack problem deﬁned in Sec. 7.6. It was shown that
using dynamic programming, the problem can be solved in time Θ(nC),
where n is the number of items and C is the knapsack capacity.
(a) Give a backtracking algorithm to solve the knapsack problem.
(b) Which technique is more eﬃcient to solve the knapsack problem:
backtracking or dynamic programming? Explain.
13.15. Give a backtracking algorithm to solve the money change problem de-
ﬁned in Exercise 7.30.
13.16. Apply the algorithm in Exercise 13.15 for the money change problem on
the instance in Exercise 7.31.
13.17. Give a backtracking algorithm to solve the assignment problem deﬁned
as follows. Given n employees to be assigned to n jobs such that the cost
of assigning the ith person to the jth job is c
i,j
, ﬁnd an assignment that
minimizes the total cost. Assume that the cost is nonnegative, that is,
c
i,j
≥ 0 for 1 ≤ i, j ≤ n.
13.18. Modify the solution of the instance of the traveling salesman problem
given in Sec. 13.5 so that it results in fewer node splittings.
13.19. Apply the branch-and-bound algorithm for the traveling salesman
problem discussed in Sec. 13.5 on the instance

∞ 5 2 10
2 ∞ 5 12
3 7 ∞ 5
8 2 4 ∞
¸
¸
¸
.
13.20. Consider again the knapsack problem deﬁned in Sec. 7.6. Use branch
and bound and a suitable lower bound to solve the instance of this prob-
lem in Example 7.6.
Bibliographic notes 369
13.21. Carry out a branch-and-bound procedure to solve the following instance
of the assignment problem deﬁned in Exercise 13.17. There are four
employees and four jobs. The cost function is represented by the ma-
trix below. In this matrix, row i corresponds to the ith employee, and
column j corresponds to the jth job.

3 5 2 4
6 7 5 3
3 7 4 5
8 5 4 6
¸
¸
¸
.
13.7 Bibliographic notes
There are several books that cover backtracking in some detail. These
include Brassard and Bratley (1988), Horowitz and Sahni (1978), Reingold,
Nievergelt and Deo (1977). It is also described in Golomb and Brumert
(1965). Techniques for analyzing its eﬃciency are given in Knuth (1975).
The recursive form of backtracking was used by Tarjan (1972) in various
graph algorithms. Branch-and-bound techniques have been successfully
used in optimization problems since the late 1950s. Many of the diverse
applications are outlined in the survey paper by Lawler and Wood (1966).
The approach to solve the traveling salesman problem in this chapter
is due to Little, Murty, Sweeney and Karel (1963). Another technique to
solve the traveling salesman problem is described in the survey paper
by Bellmore and Nemhauser (1968).
370
Chapter 14
Randomized Algorithms
14.1 Introduction
In this chapter we discuss one form of algorithm design in which we relax
the condition that an algorithm must solve the problem correctly for all
possible inputs, and demand that its possible incorrectness is something
that can safely be ignored due, say, to its very low likelihood of occur-
rence. Also, we will not demand that the output of an algorithm must be
the same in every run on a particular input. We will be concerned with
those algorithms that in the course of their execution can toss a fair coin,
yielding truly random outcomes. The consequences of adding this element
of randomness turn out to be surprising. Rather than producing unpre-
dictable results, the randomness introduced will be shown to be extremely
useful and capable of yielding fast solutions to problems that have only very
ineﬃcient deterministic algorithms.
A randomized algorithm can be deﬁned as one that receives, in addition
to its input, a stream of random bits that it can use in the course of its
action for the purpose of making random choices. A randomized algorithm
may give diﬀerent results when applied to the same input in diﬀerent runs.
It follows that the execution time of a randomized algorithm may vary
from one run to another when applied to the same input. By now, it
is recognized that, in a wide range of applications, randomization is an
extremely important tool for the construction of algorithms. There are two
main advantages that randomized algorithms often have. First, often the
execution time or space requirement of a randomized algorithm is smaller
than that of the best deterministic algorithm that we know of for the same
371
372 Randomized Algorithms
problem. Second, if we look at the various randomized algorithms that have
been invented so far, we ﬁnd that invariably they are extremely simple
to comprehend and implement. The following is a simple example of a
randomized algorithm.
Example 14.1 Suppose we have a polynomial expression in n variables, say
f(x
1
, x
2
, . . . , x
n
), and we wish to check whether or not f is identically zero. To
do this analytically could be a horrendous job. Suppose, instead, we generate a
random n-vector (r
1
, r
2
, . . . , r
n
) of real numbers and evaluate f(r
1
, r
2
, . . . , r
n
). If
f(r
1
, r
2
, . . . , r
n
) ,= 0, we know that f ,= 0. If f(r
1
, r
2
, . . . , r
n
) = 0, then either f is
identically zero or we have been extremely lucky in our choice of (r
1
, r
2
, . . . , r
n
).
If we repeat this several times and keep on getting f = 0, then we conclude that
f is identically zero. The probability that we have made an error is negligible.
In some deterministic algorithms, especially those that exhibit good
average running time, the mere introduction of randomization suﬃces to
convert a simple and na¨ıve algorithm with bad worst case behavior into
a randomized algorithm that performs well with high probability on every
possible input. This will be apparent when we study randomized algorithms
for sorting and selection in Sec. 14.3 and Sec. 14.4.
14.2 Las Vegas and Monte Carlo Algorithms
Randomized algorithms can be classiﬁed into two categories. The ﬁrst cate-
gory is referred to as Las Vegas algorithms. It constitutes those randomized
algorithms that always give a correct answer, or do not give an answer at
all. This is to be contrasted with the other category of randomized algo-
rithms, which are referred to as Monte Carlo algorithms. A Monte Carlo
algorithm always gives an answer, but may occasionally produce an answer
that is incorrect. However, the probability of producing an incorrect an-
swer can be made arbitrarily small by running the algorithm repeatedly
with independent random choices in each run.
To be able to generally discuss the computational complexity of a ran-
domized algorithm, it is useful to ﬁrst introduce some criteria for evaluating
the performance of algorithms. Let A be an algorithm. If A is determinis-
tic, then one measure of the time complexity of the algorithm is its average
running time: The average time taken by A when for each value of n, each
input of size n is considered equally likely. That is, a uniform distribution
Randomized Quicksort 373
on all its inputs is assumed (see Sec. 1.12.2). This may be misleading, as
the input distribution may not be uniform. If A is a randomized algorithm,
then its running time on a ﬁxed instance I of size n may vary from one
execution to another. Therefore, a more natural measure of performance
is the expected running time of A on a ﬁxed instance I. This is the mean
time taken by algorithm A to solve the instance I over and over. Thus, it
is natural to talk about the worst case expected time and the average case
expected time (see Exercises 14.3 and 14.4).
14.3 Randomized Quicksort
This is, perhaps, one of the most popular randomized algorithms. Consider
Algorithmquicksort which was presented in Sec. 6.6. We have shown that
the algorithm’s running time is Θ(nlog n) on the average, provided that all
permutations of the input elements are equally likely. This, however, is not
the case in many practical applications. We have also shown that, ironically,
if the input is already sorted, then its running time is Θ(n
2
). This is also
the case if the input is almost sorted. Consider, for instance, an application
that updates a large sorted ﬁle by appending a small number of elements
to the original sorted ﬁle and then using Algorithm quicksort to sort it
afresh. In this case, the smaller the number of added elements the closer
the running time is to Θ(n
2
).
One approach to circumvent this problem and guarantee a running time
of Θ(nlog n) on the average is to introduce a preprocessing step whose sole
purpose is to permute the elements to be sorted randomly. This preprocess-
ing step can be performed in Θ(n) time (Exercise 14.5). Another simpler
approach which leads to the same eﬀect is to introduce an element of ran-
domness into the algorithm. This can be done by selecting the pivot on
which to split the elements randomly. The result of choosing the pivot
randomly is to relax the assumption that all permutations of the input el-
ements are equally likely. Modifying the original Algorithm quicksort by
introducing this step results in Algorithm randomizedquicksort shown
below. The new algorithm simply chooses uniformly at random an index
v in the interval [low..high] and interchanges A[v] with A[low]. This is
because Algorithm split uses A[low] as the pivot (see Sec. 6.6.1). The
algorithm then continues as in the original quicksort algorithm. Here,
the function random(low, high) returns a number between low and high
374 Randomized Algorithms
randomly. It is important to note that any number between low and high
is generated with equal probability of 1/(high −low + 1).
Algorithm 14.1 randomizedquicksort
Input: An array A[1..n] of n elements.
Output: The elements in A sorted in nondecreasing order.
1. rquicksort(1, n)
Procedure rquicksort(low, high)
1. if low < high then
2. v←random(low, high)
3. interchange A[low] and A[v]
4. split(A[low..high], w) ¦w is the new position of the pivot¦
5. rquicksort(low, w −1)
6. rquicksort(w + 1, high)
7. end if
The analysis of the running time of Algorithm randomizedquicksort
is almost the same as the original one, except that now we don’t have to
assume that all permutations of the input elements are equally likely. The
running time of the algorithm is still Θ(n
2
) in the worst case, but this is
not attributed to the input form; if it does happen, then it is because of the
unlucky random pivots chosen by the random number generator, which is
very unlikely. In fact, no permutation of the input can elicit its worst case
behavior, and the algorithm’s expected running time is Θ(nlog n).
14.4 Randomized Selection
Consider Algorithm select, which was presented in Sec. 6.5. We have
shown that the algorithm’s running time is Θ(n) with a large multiplica-
tive constant that makes the algorithm impractical, especially for small and
moderate values of n. In this section, we present a randomized Las Vegas
algorithm for selection that is both simple and fast. Its expected running
time is Θ(n) with a small multiplicative constant. As in Algorithm ran-
domizedquicksort, its running time is Θ(n
2
) in the worst case. As in
Algorithm randomizedquicksort too, this worst case bound is indepen-
dent of the instance being solved by the algorithm; it is only a result of an
unlucky sequence of choices made by the random number generator, which
Randomized Selection 375
is very unlikely. The algorithm behaves like the binary search algorithm in
the sense that it keeps discarding portions of the input until one element is
left, which is the desired kth smallest element. A precise description of the
algorithm is given in Algorithm randomizedselect.
Algorithm 14.2 randomizedselect
Input: An array A[1..n] of n elements and an integer k, 1 ≤ k ≤ n.
Output: The kth smallest element in A.
1. rselect(A, 1, n, k)
Procedure rselect(A, low, high, k)
1. v←random(low, high)
2. x←A[v].
3. Partition A[low..high] into three arrays:
A
1
= ¦a [ a < x¦
A
2
= ¦a [ a = x¦
A
3
= ¦a [ a > x¦
4. case
[A
1
[ ≥ k: return rselect(A
1
, 1, [A
1
[, k)
[A
1
[ +[A
2
[ ≥ k: return x
[A
1
[ +[A
2
[ < k: return rselect(A
3
, 1, [A
3
[, k −[A
1
[ −[A
2
[)
5. end case
In what follows we investigate the running time of Algorithm random-
izedselect. Assume without loss of generality that the elements in A are
distinct. In the worst case, the random number generator selects the jth
element in the jth iteration, and thus only one element is discarded. Since
the number of comparisons needed to partition A into A
1
, A
2
and A
3
is
exactly n, the total number of element comparisons done by the algorithm
in the worst case is
n + (n −1) +. . . + 1 =
n(n + 1)
2
= Θ(n
2
).
Now, we prove by induction that the expected number of element com-
parisons done by the algorithm is less than 4n. Let C(n) be the expected
number of element comparisons performed by the algorithm on a sequence
of n elements. Since v, which is chosen randomly, may assume any of the
integers 1, 2, . . . , n with equal probability, we have two cases to consider
according to whether v < k or v > k. If v < k, the number of remain-
ing elements is n − v, and if v > k, the number of remaining elements is
376 Randomized Algorithms
v − 1. Thus, the expected number of element comparisons performed by
the algorithm is
C(n) = n +
1
n

= 4n −2
< 4n.
Since each element is inspected at least once, C(n) ≥ n. Thus, we have
the following theorem:
Theorem 14.1 The expected number of element comparisons performed
by Algorithm randomizedselect on input of size n is less than 4n. Its
expected running time is Θ(n).
14.5 Testing String Equality
In this section we outline an example of how randomization can be employed
to cut down drastically on the communication cost. Suppose that two
parties A and B can communicate over a communication channel, which
we will assume to be very reliable. A has a very long string x and B has a
very long string y, and they want to determine whether x = y. Obviously,
A can send x to B, who in turn can immediately test whether x = y. But
this method would be extremely expensive, in view of the cost of using
the channel. Another alternative would be for A to derive from x a much
shorter string that could serve as a “ﬁngerprint” of x and send it to B.
B then would use the same derivation to obtain a ﬁngerprint for y, and
then compare the two ﬁngerprints. If they are equal, then B would assume
that x = y; otherwise he would conclude that x = y. B then notiﬁes A of
the outcome of the test. This method requires the transmission of a much
shorter string across the channel. For a string w, let I(w) be the integer
represented by the bit string w. One method of ﬁngerprinting is to choose
a prime number p and then use the ﬁngerprint function
I
p
(x) = I(x) (mod p).
If p is not too large, then the ﬁngerprint I
p
(x) can be sent as a short string.
The number of bits to be transmitted is thus O(log p). If I
p
(x) = I
p
(y),
then obviously x = y. However, the converse is not true. That is, if
378 Randomized Algorithms
I
p
(x) = I
p
(y), then it is not necessarily the case that x = y. We refer
to this phenomenon as a false match. In general, a false match occurs if
x = y, but I
p
(x) = I
p
(y), i.e., p divides I(x) − I(y). We will later bound
the probability of a false match.
The weakness of this method is that, for ﬁxed p, there are certain pairs
of strings x and y on which the method will always fail. We get around the
problem of the existence of these pairs x and y by choosing p at random
every time the equality of two strings is to be checked, rather than agreeing
on p in advance. Moreover, choosing p at random allows for resending
another ﬁngerprint, and thus increasing the conﬁdence in the case x = y.
This method is described in Algorithm stringequalitytest shown below
(The value of M will be determined later).
Algorithm 14.3 stringequalitytest
1. A chooses p at random from the set of primes less than M.
2. A sends p and I
p
(x) to B.
3. B checks whether I
p
(x) = I
p
(y) and conﬁrms the equality or inequal-
ity of the two strings x and y.
Now, we compute the probability of a false match. Let n be the number
of bits in the binary representation of x. Of course, n is also equal to the
number of bits in the binary representation of y; otherwise the problem
is trivial. Let π(n) be the number of distinct primes less than n. It is
well-known that π(n) is asymptotic to n/ ln n. It is also known that if
k < 2
n
, then, except when n is very small, the number of distinct primes
that divide k is less than π(n). Since failure can occur only in the case of a
false match, i.e., x = y but I
p
(x) = I
p
(y), this is only possible if p divides
I(x) −I(y). Hence, the probability of failure for two n-bit strings x and y
is
[¦p [ p is a prime < 2
n
and p divides I(x) −I(y)¦[
π(M)
≤
π(n)
π(M)
.
If we choose M = 2n
2
, we obtain
Pr [failure] ≤
π(n)
π(M)
≈
n/ ln n
2n
2
/ ln n
2
=
1
n
.
Furthermore, if we repeat the algorithm k times, each time selecting
a prime number less than M at random, then the probability becomes at
Pattern matching 379
most (1/n)
k
. If, for example, we set k = log log n|, then the probability
of failure becomes
Pr [failure] ≤
1
n
log log n
.
Example 14.2 Suppose that x and y are one million bits each, i.e., n =
1, 000, 000. Then, M = 2 10
12
= 2
40.8631
. In this case, the number of bits
required to transmit p is at most ]log M + 1 = 40 + 1 = 41. The number
of bits required to transmit the ﬁngerprint of x is at most ]log(p −1) + 1 ≤
]log M +1 = 41. Thus, the total number of bits transmitted is at most 82. The
probability of failure in one transmission is at most 1/n = 1/1, 000, 000. Since
]log log n = 5, repeating the algorithm ﬁve times reduces the probability of false
match to n
−log log n
= (10
6
)
−5
= 10
−30
, which is negligible.
14.6 Pattern matching
Now we apply the same idea of ﬁngerprinting described in Sec. 14.5 to a
classical problem in computer science: pattern matching. Given a string
of text X = x
1
x
2
. . . x
n
and a pattern Y = y
1
y
2
. . . y
m
, where m ≤ n,
determine whether or not the pattern appears in the text. Without loss
of generality, we will assume that the text alphabet is Σ = ¦0, 1¦. The
most straightforward method for solving this problem is simply to move the
pattern across the entire text, and in every position compare the pattern
with the portion of the text of length m. This brute-force method leads
to an O(mn) running time in the worst case. There are, however, more
complicated deterministic algorithms whose running time is O(n +m).
Here we will present a simple and eﬃcient Monte Carlo algorithm that
also achieves a running time of O(n + m). We will convert it later into
a Las Vegas algorithm having the same time complexity. The algorithm
follows the same brute-force algorithm of sliding the pattern Y across the
text X, but instead of comparing the pattern with each block X(j) =
x
j
x
j+1
. . . x
j+m−1
, we will compare the ﬁngerprint I
p
(Y ) of the pattern
with the ﬁngerprints I
p
(X(j)) of the blocks of text. The O(n) ﬁngerprints
of the text are fortunately easy to compute. The key observation is that
when we shift from one block of text to the next, the ﬁngerprint of the
new block X(j + 1) can easily be computed from the ﬁngerprint of X(j).
380 Randomized Algorithms
Speciﬁcally,
I
p
(X(j + 1)) = (2I
p
(X(j)) −2
m
x
j
+x
j+m
) (mod p).
If we let W
p
= 2
m
(mod p), then we have the recurrence
I
p
(X(j + 1)) = (2I
p
(X(j)) −W
p
x
j
+x
j+m
) (mod p). (14.2)
The pattern matching algorithm is shown as Algorithm pattern-
matching (The value of M will be determined later).
Algorithm 14.4 patternmatching
Input: A string of text X and a pattern Y of length n and m, respectively.
Output: The ﬁrst position of Y in X if Y occurs in X; otherwise 0.
1. Choose p at random from the set of primes less than M.
2. j ←1
3. Compute W
p
= 2
m
(mod p), I
p
(Y ) and I
p
(X
j
)
4. while j ≤ n −m+ 1
5. if I
p
(X
j
) = I
p
(Y ) then return j ¦A match is found (proba-
bly)¦
6. Compute I
p
(X
j
) using Eqn. 14.2.
7. j ←j + 1
8. end while
9. return 0 ¦Y does not occur in X (deﬁnitely)¦
The computation of each of W
p
, I
p
(Y ) and I
p
(X(1)) costs O(m) time.
When implementing the computation of I
p
(X(j + 1)) from I
p
(X(j)), we
do not need to use the more expensive operations of multiplication and
division; only a constant number of additions and subtractions is needed.
Thus, the computation of each I
p
(X(j)), for 2 ≤ j ≤ n − m + 1, costs
only O(1) time for a total of O(n) time. Hence, the running time is O(n +
m). The above analysis is valid under the uniform-cost RAM model of
computation. If we use the more realistic logarithmic-cost RAM model of
computation, then the time complexity is increased by a factor of log p.
Now we analyze the frequency with which this algorithm will fail. A
false match will occur only if for some j we have
Y = X(j) but I
p
(Y ) = I
p
(X(j)).
Random Sampling 381
This is only possible if the chosen prime number p divides
¸
{j [ Y =X(j)}
[I(Y ) −I(X(j))[.
This product cannot exceed (2
m
)
n
= 2
mn
(Sec. 14.5), and hence the number
of primes that divide it cannot exceed π(mn). If we choose M = 2mn
2
,
then the probability of a false match cannot exceed
π(mn)
π(M)
≈
mn/ ln(mn)
2mn
2
/ ln(mn
2
)
=
ln(mn
2
)
2nln(mn)
<
ln(mn)
2
2nln(mn)
=
1
n
.
It is interesting to note that, according to the above derivation, the
probability of failure depends only on the length of the text, and the length
of the pattern has no eﬀect on this probability. Note also that in the case
when m = n, the problem reduces to that of testing the equality of two
strings of equal length discussed in (Sec. 14.5), and that the probability of
failure is identical to the one derived for that problem.
To convert the algorithm into a Las Vegas algorithm is easy. Whenever
the two ﬁngerprints I
p
(Y ) and I
p
(X(j)) match, the two strings are tested
for equality. The expected time complexity of this Las Vegas algorithm
becomes
O(n +m)

1 −
1
n

+mn

1
n

= O(n +m).
Thus, we end with an eﬃcient pattern matching algorithm that always
gives the correct result.
14.7 Random Sampling
Consider the problem of selecting a sample of m elements randomly from
a set of n elements, where m < n. For simplicity, we will assume that the
elements are positive integers between 1 and n. In this section, we present
a simple Las Vegas algorithm that runs in time Θ(n).
Consider the following selection method. First mark all the n elements
as unselected. Next, repeat the following step until exactly m elements
have been selected. Generate a random number r between 1 and n. If
r is marked unselected, then mark it selected and add it to the sample.
This method is described more precisely in Algorithm randomsampling.
A disadvantage of this algorithm is that its space complexity is Θ(n), as
382 Randomized Algorithms
it needs an array to mark each integer. If n is too large compared to m
(e.g. n > m
2
), the algorithm can easily be modiﬁed to eliminate the need
for this array (see Exercise 14.16).
Algorithm 14.5 randomsampling
Input: Two positive integers m, n with m < n.
Output: An array A[1..m] of m distinct positive integers selected
randomly from the set ¦1, 2, . . . , n¦.
1. comment: S[1..n] is a boolean array indicating
whether an integer has been selected.
2. for i ←1 to n
3. S[i] ←false
4. end for
5. k←0
6. while k < m
7. r ←random(1, n)
8. if not S[r] then
9. k←k + 1
10. A[k] ←r
11. S[r] ←true
12. end if
13. end while
Clearly, the smaller the diﬀerence between m and n the larger the run-
ning time. For example, if n = 1000 and m = 990, then the algorithm
will spend too much time in order to generate the last integers in the
sample, e.g. the 990th integer. To circumvent this problem, we may select
10 integers randomly, discard them and keep the remaining 990 integers as
the desired sample. Therefore, we will assume that m ≤ n/2, since other-
wise we may select n − m integers randomly, discard them and keep the
remaining integers as our sample.
Before we analyze the running time of the algorithm, we need the fol-
lowing deﬁnition:
Deﬁnition 14.1 Let c be a coin whose probability of “heads” is p, and
let q = 1 − p. Suppose we ﬂip c repeatedly until “heads” appears for the
ﬁrst time. Let X be the random variable denoting the total number of
ﬂips until “heads” appears for the ﬁrst time. Then X is said to have the
Random Sampling 383
geometric distribution with probability
Pr [X = k] =

pq
k−1
if k ≥ 1
0 if k < 1.
The expected value of X is
E(X) =
∞
¸
k=1
kpq
k−1
=
1
p
. (14.3)
Let p
k
be the probability of generating an unselected integer given that
k −1 numbers have already been selected, where 1 ≤ k ≤ m. Clearly,
p
k
=
n −k + 1
n
.
If X
k
, 1 ≤ k ≤ m, is the random variable denoting the number of integers
generated in order to select the kth integer, then X
k
has the geometric
distribution with expected value
E(X
k
) =
1
p
k
=
n
n −k + 1
.
Let Y be the random variable denoting the total number of integers gener-
ated in order to select the m out of n integers. By linearity of expectation,
we have
E(Y ) = E(X
1
) +E(X
2
) +. . . +E(X
m
).
Hence,
E(Y ) =
m
¸
k=1
E(X
k
)
=
m
¸
k=1
n
n −k + 1
= n
n
¸
k=1
1
n −k + 1
−n
n
¸
k=m+1
1
n −k + 1
= n
n
¸
k=1
1
k
−n
n−m
¸
k=1
1
k
.
384 Randomized Algorithms
By Eq. 2.16 on page 80,
n
¸
j=1
1
k
≤ ln n + 1 and
n−m
¸
k=1
1
k
≥ ln(n −m+ 1).
Hence,
E(Y ) ≤ n(ln n + 1 −ln(n −m+ 1))
≈ n(ln n + 1 −ln(n −m))
≤ n(ln n + 1 −ln(n/2)) since m ≤ n/2
= n(ln 2 + 1)
= n ln 2e
≈ 1.69 n.
Since T(n), the expected running time of the algorithm, is proportional
to E(Y ), we have as a result T(n) = Θ(n).
14.8 Primality Testing
In this section, we study a well-known Monte Carlo algorithm for testing
whether a given positive integer n is prime. The obvious method of repeat-
edly dividing by the numbers from 2 to
√
n| is extremely ineﬃcient, as it
leads to exponential time complexity in the input size (see Example 1.16).
This method is appropriate only for small numbers, and its only advantage
is that it outputs a divisor of n if it is composite. It turns out that factoring
an integer is a much harder problem than merely testing whether it is prime
or composite.
In primality tests, we use the idea of ﬁnding a “witness”, which is a
proof that a number is composite. Obviously, ﬁnding a divisor of n is a
proof that it is composite. However, such witnesses are very rare. Indeed,
if we take a number n that is fairly large, the number of its prime divisors
is very small compared to the number of integers between 1 and n. It is
well-known that if n < 2
k
, then, except when k is very small, the number
of distinct primes that divide n is less than π(k) ≈ k/ ln k.
This motivates the search for another type of witness. Before discussing
the alternate witness, we will dispose of an algorithm for an operation that
will be used throughout this section. Let a, m and n be positive integers
Primality Testing 385
with m ≤ n. We need the operation of raising a to the mth power and reduc-
ing the result modulo n. Algorithm expmod below computes a
m
(mod n).
It is similar to the exponentiation algorithm presented in Sec. 5.4. Notice
that we reduce modulo n after each squaring or multiplication rather than
ﬁrst computing a
m
and reducing modulo n once at the end. A call to this
algorithm is of the form expmod(a, m, n).
Algorithm 14.6 expmod
Input: Positive integers a, m and n with m ≤ n.
Output: a
m
(mod n).
1. Let the binary digits of m be b
k
, b
k−1
, . . . , b
0
.
2. c←1
3. for j ←k downto 0
4. c←c
2
(mod n)
5. if b
j
= 1 then c←ac (mod n)
6. end for
7. return c
It is easy to see that the running time of Algorithm expmod is
Θ(log m) = O(log n), if we want to charge one unit of time per one multipli-
cation. However, since we are dealing here with arbitrarily large integers,
we will count the exact number of bit multiplications performed by the al-
gorithm. If we use the obvious way of multiplying two integers, then each
multiplication costs O(log
2
n). Thus, the overall running time of Algorithm
expmod is O(log
3
n).
Now, we present a series of primality tests all of which are based on
Fermat’s theorem:
Theorem 14.2 If n is prime, then for all a ≡ 0 (mod n) we have
a
n−1
≡ 1 (mod n).
Consider Algorithm ptest1. By Fermat’s theorem, if Algorithm
ptest1 returns composite, then we are sure that n is composite. The
ancient Chinese conjectured that a natural number n must be prime if it
satisﬁes the congruence 2
n
≡ 2 (mod n). The question remained open until
1819, when Sarrus showed that 2
340
≡ 1 (mod 341), and yet 341 = 11 31
is composite. Some other composite numbers that satisfy the congruence
386 Randomized Algorithms
2
n−1
≡ 1 (mod n) are 561, 645, 1105, 1387, 1729 and 1905. Thus, if Algo-
rithm ptest1 returns prime, then n may or may not be prime.
Algorithm 14.7 ptest1
Input: A positive odd integer n ≥ 5.
Output: prime if n is prime; otherwise composite.
1. if expmod(2, n−1, n) ≡ 1 (mod n) then return prime ¦probably¦
2. else return composite ¦deﬁnitely¦
Surprisingly, this simple test gives an erroneous result very rarely. For
example, for all composite numbers between 4 and 2000, the algorithm
returns prime only for the numbers 341, 561, 645, 1105, 1387, 1729 and
1905. Moreover, there are only 78 values of n less than 100,000 for which
the test errs, the largest of which is 93961 = 7 31 433.
It turns out, however, that for many composite numbers n, there exist
integers a for which a
n−1
≡ 1 (mod n). In other words, the converse of
Fermat’s theorem is not true (we have already proven this for a = 2).
Indeed, there are composite integers n known as Carmichael numbers that
satisfy Fermat’s theorem for all positive integers a that are relatively prime
to n. The smallest Carmichael numbers are 561 = 3 11 17, 1105 =
5 13 17, 1729 = 7 13 19 and 2465 = 5 17 29. Carmichael
numbers are very rare; there are, for example, only 255 of them less than
10
8
. When a composite number n satisﬁes Fermat’s theorem relative to
base a, n is called a base-a pseudoprime. Thus, Algorithm ptest1 returns
prime whenever n is prime or base-2 pseudoprime.
One way to improve on the performance of Algorithm ptest1 is to
choose the base randomly between 2 and n − 2. This yields Algorithm
ptest2.
As in Algorithm ptest1, Algorithm ptest2 errs only if n is a base-
a pseudoprime. For example, 91 = 7 13 is base-3 pseudoprime since
3
90
≡ 1 (mod 91).
Let Z
∗
n
be the set of positive integers less than n that are relatively
prime to n. It is well-known that Z
∗
n
forms a group under the operation of
multiplication modulo n. Deﬁne
F
n
= ¦a ∈ Z
∗
n
[ a
n−1
≡ 1 (mod n)¦.
Primality Testing 387
Algorithm 14.8 ptest2
Input: A positive odd integer n ≥ 5.
Output: prime if n is prime; otherwise composite.
1. a←random(2, n −2)
2. if expmod(a, n−1, n) ≡ 1 (mod n) then return prime ¦probably¦
3. else return composite ¦deﬁnitely¦
If n is prime or a Carmichael number, then F
n
= Z
∗
n
. So, suppose n is not a
Carmichael number or a prime number. Then, F
n
= Z
∗
n
. It is easy to verify
that F
n
under the operation of multiplication modulo n forms a group that
is a proper subgroup of Z
∗
n
. Consequently, the order of F
n
divides the order
of Z
∗
n
, that is, [F
n
[ divides [Z
∗
n
[. It follows that the number of elements in
F
n
is at most half the number of elements in Z
∗
n
. This proves the following
lemma:
Lemma 14.1 If n is not a Carmichael number, then Algorithm ptest2
will detect the compositeness of n with probability at least 1/2.
Unfortunately, it has recently been shown that there are, in fact, in-
ﬁnitely many Carmichael numbers. In the remainder of this section, we
describe a more powerful randomized primality test algorithm that circum-
vents the diﬃculty that arises as a result of the existence of inﬁnitely many
Carmichael numbers. The algorithm has the property that if n is com-
posite, then the probability that this is discovered is at least
1
2
. In other
words, the probability that it will err is at most
1
2
. Thus, by repeating the
test k times, the probability that it will err is at most 2
−k
. The algorithm,
which we will call ptest3, is based on the following reasoning. Let n ≥ 5
be an odd prime. Write n −1 = 2
q
m (q ≥ 1 since n −1 is even). Then, by
Fermat’s theorem, the sequence
a
m
(mod n), a
2m
(mod n), a
4m
(mod n), . . . , a
2
q
m
(mod n)
must end with 1, and the value just preceding the ﬁrst appearance of 1 will
be n −1. This is because the only solutions to x
2
≡ 1 (mod n) are x = ±1
when n is prime, since in this case Z
∗
n
is a ﬁeld. This reasoning leads to
Algorithm ptest3.
388 Randomized Algorithms
Algorithm 14.9 ptest3
Input: A positive odd integer n ≥ 5.
Output: prime if n is prime; otherwise composite.
1. q ←0; m←n −1
2. repeat ¦ﬁnd q and m¦
3. m←m/2
4. q ←q + 1
5. until m is odd
6. a←random(2, n −2)
7. x←expmod(a, m, n)
8. if x = 1 then return prime ¦probably¦
9. for j ←0 to q −1
10. if x ≡ −1 (mod n) then return prime ¦probably¦
11. x←x
2
(mod n)
12. end for
13. return composite ¦deﬁnitely¦
Theorem 14.3 If Algorithmptest3 returns “composite”, then n is com-
posite.
Proof. Suppose that Algorithm ptest3 returns “composite”, but n is an
odd prime. We claim that a
2
j
m
≡ 1 (mod n) for j = q, q − 1, . . . , 0. If so,
then setting j = 0 yields a
m
≡ 1 (mod n), which means that the algorithm
must have returned “prime” by Step 8, a contradiction to the outcome of
the algorithm. This contradiction establishes the theorem. Now, we prove
our claim. By Fermat’s theorem, since n is prime, the statement is true for
j = q. Suppose it is true for some j, 1 ≤ j ≤ q. Then it is also true for
j −1 also because
(a
2
j−1
m
)
2
= a
2
j
m
≡ 1 (mod n)
implies that the quantity being squared is ±1. Indeed, the equation x
2
= 1
in Z
∗
n
has only the solution x = ±1, since in this case Z
∗
n
is a ﬁeld. But −1
is ruled out by the outcome of the algorithm since it must have executed
Step 13. Consequently, a
2
j−1
m
= 1. This completes the proof of the claim.
It follows that n is composite. ⁄
Note that the contrapositive statement of the above theorem is: If n
is prime, then Algorithm ptest3 returns “prime”, which means that the
algorithm will never err if n is prime.
Primality Testing 389
Obviously, Algorithm ptest3 is as good as Algorithm ptest2 in deal-
ing with non-Carmichael numbers. It can be shown, although we will not
pursue it here, that the probability that Algorithm ptest3 errs when pre-
sented with a Carmichael number is at most 1/2. So, the probability that
it will err on any composite number is at most 1/2. Thus, by repeating
the test k times, the probability that it will err is at most 2
−k
. If we set
k = log n|, the probability of failure becomes 2
−log n
≤ 1/n. In other
words, the algorithm will give the correct answer with probability at least
1 −1/n, which is negligible when n is suﬃciently large. This results in our
ﬁnal algorithm, which we will call primalitytest.
Algorithm 14.10 primalitytest
Input: A positive odd integer n ≥ 5.
Output: prime if n is prime; otherwise composite.
1. q ←0; m←n −1; k←]log n
2. repeat ¦ﬁnd q and m¦
3. m←m/2
4. q ←q + 1
5. until m is odd
6. for i ←1 to k
7. a←random(2, n −2)
8. x←expmod(a, m, n)
9. if x = 1 then return prime ¦probably¦
10. for j ←0 to q −1
11. if x ≡ −1 (mod n) then return prime ¦probably¦
12. x←x
2
(mod n)
13. end for
14. end for
15. return composite ¦deﬁnitely¦
We compute the running time of Algorithm primalitytest as follows
(assuming that a random integer can be generated in O(1) time). The
repeat loop costs O(log n) time. We have shown before that the cost of
Step 8 is Θ(log
3
n). The number of squarings in the inner for loop is
O(log n), and each costs O(log
2
n) for a total of O(log
3
n) time. Since
this is repeated k = log n| times, the time complexity of the algorithm is
O(log
4
n). The following theorem summarizes the main result.
Theorem 14.4 In time O(log
4
n), Algorithm primalitytest behaves
as follows when presented with an odd integer n ≥ 5:
390 Randomized Algorithms
(1) If n is prime, then it outputs prime.
(2) If n is composite, then it outputs composite with probability at
least 1 −1/n.
14.9 Exercises
14.1. Let p
1
, p
2
and p
3
be three polynomials of degrees n, n and 2n, re-
spectively. Give a randomized algorithm to test whether p
3
(x) =
p
1
(x) p
2
(x).
14.2. Let n be a positive integer. Design an eﬃcient randomized algorithm
that generates a random permutation of the integers 1, 2, . . . , n. Assume
that you have access to a fair coin. Analyze the time complexity of your
algorithm.
14.3. Carefully deﬁne what it means for a randomized algorithm to run in
expected time T(n) in the worst case (see Sec. 1.12).
14.4. Carefully deﬁne what it means for a randomized algorithm to run in
expected time T(n) on the average (see Sec. 1.12).
14.5. In the discussion of Algorithm randomizedquicksort, it was stated
that one possibility to obtain a Θ(nlog n) expected time for Algorithm
quicksort is by permuting the input elements so that their order be-
comes random. Describe an O(n) time algorithm to randomly permute
the input array before processing it by Algorithm quicksort.
14.6. Show that Eq. 14.1 is maximum when k = ]n/2.
14.7. Consider the following modiﬁcation of Algorithm binarysearch (see
Sec. 1.3). Instead of halving the search interval in each iteration, select
one of the remaining positions at random. Assume that every position
between low and high is equally likely to be chosen by the algorithm.
Compare the performance of this algorithm with that of Algorithm bi-
narysearch.
14.8. Let A be a Monte Carlo algorithm whose expected running time is at
most T(n) and gives a correct solution with probability p(n). Suppose
the correctness of any solution of the algorithm can always be veriﬁed in
time T

(n). Show that A can be converted into a Las Vegas algorithm
A

for the same problem that runs in expected time at most (T(n) +
T

(n))/p(n).
14.9. Suppose that a Monte Carlo algorithm gives the correct solution with
probability at least 1−
1
, regardless of the input. How many executions
of the same algorithm are necessary in order to raise the probability to
at least 1 −
2
, where 0 <
2
<
1
< 1.
Exercises 391
14.10. Let L = x
1
, x
2
, . . . , x
n
be a sequence of elements that contains exactly
k occurrences of the element x (1 ≤ k ≤ n). We want to ﬁnd one j
such that x
j
= x. Consider repeating the following procedure until x is
found. Generate a random number i between 1 and n and check whether
x
i
= x. Which method is faster, on the average, this method or linear
search? Explain.
14.11. Let S = ¦x
1
, x
2
, . . . , x
n
¦ be a set of n positive integers. We want to
ﬁnd an element x that is in the upper half when S is sorted, or in other
words greater than the median. Give a randomized algorithm to solve
this problem with probability 1 − 1/n. Compare this algorithm with a
deterministic algorithm for large values of n, say more than a million.
14.12. Let L be a list of n elements that contains a majority element (see
Sec. 5.7). Give a randomized algorithm that ﬁnds the majority element
with probability 1 − , for a given > 0. Is randomization suitable for
this problem in view of the fact that there is an O(n) time algorithm to
solve it?
14.13. Let A, B and C be three n n matrices. Give a Θ(n
2
) time algorithm
to test whether AB = C. The algorithm is such that if AB = C,
then it returns true. What is the probability that it returns true when
AB ,= C? (Hint: Let x be a vector of n random entries. Perform the
test A(BX) = CX).
14.14. Let A and B be two nn matrices. Give a Θ(n
2
) time algorithm to test
whether A = B
−1
.
14.15. Consider the sampling problem in Sec. 14.7. Suppose we perform one
pass over the n integers and choose each one with probability m/n. Show
that the size of the resulting sample has a large variance, and hence its
size may be much smaller or larger than m.
14.16. Modify Algorithm randomsampling in Sec. 14.7 to eliminate the need
for the boolean array S[1..n]. Assume that n is too large compared to
m, say n > m
2
.
14.17. What are the time and space complexities of the algorithm in Exer-
cise 14.16?
14.18. Prove Eq. 14.3.
14.19. Consider F
n
as deﬁned on page 386. Suppose that n is neither a
Carmichael number nor a prime number. Show that F
n
under the opera-
tion of multiplication modulo n forms a group that is a proper subgroup
of Z
∗
n
.
14.20. Let G = (V, E) be a connected undirected graph with n vertices. A cut
in G is a set of edges whose removal disconnects G. A min-cut in G is a
cut with the minimum number of edges. A straightforward deterministic
approach to solve this problem is to apply the max-ﬂow min-cut theorem
392 Randomized Algorithms
(Theorem 16.1). Give a simple Monte Carlo algorithm that ﬁnds a min-
cut with reasonable probability.
14.21. A multigraph is a graph in which multiple edges are allowed between
pairs of vertices. Show that the number of distinct min-cuts in a multi-
graph with n vertices is at most n(n −1)/2 (see Exercise 14.20).
14.22. Show that if there is a group of 23 people, then the probability that two
of them will have the same birthday is at least 1/2.
14.10 Bibliographic notes
The real start of randomized algorithms was with the publication of Ra-
bin’s paper “Probabilistic algorithms” (Rabin (1976)). In this paper, two
eﬃcient randomized algorithms were presented: one for the closest pair
problem and the other for primality testing. The probabilistic algorithm
of Solovay and Strassen (1977, 1978), also for primality testing, is another
celebrated result. Motwani and Raghavan (1995) is a comprehensive book
on randomized algorithms. Some good surveys in this ﬁeld include Karp
(1991), Welsh (1983) and Gupta, Smolka and Bhaskar (1994). Randomized
quicksort is based on Hoare (1962). The randomized selection algorithm is
due to Hoare (1961). The pattern matching algorithm is based on Karp
and Rabin (1987).
Chapter 15
Approximation Algorithms
15.1 Introduction
There are many hard combinatorial optimization problems that cannot be
solved eﬃciently using backtracking or randomization. An alternative in
this case for tackling some of these problems is to devise an approximation
algorithm, given that we will be content with a “reasonable” solution that
approximates an optimal solution. Associated with each approximation
algorithm, there is a performance bound that guarantees that the solution
to a given instance will not be far away from the neighborhood of the exact
solution. A marking characteristic of (most of) approximation algorithms
is that they are fast, as they are mostly greedy heuristics. As stated in
Chapter 8, the proof of correctness of a greedy algorithm may be complex.
In general, the better the performance bound the harder it becomes to prove
the correctness of an approximation algorithm. This will be evident when
we study some approximation algorithms. One should not be optimistic,
however, about ﬁnding an eﬃcient approximation algorithm, as there are
hard problems for which even the existence of a “reasonable” approximation
algorithm is unlikely unless NP = P.
15.2 Basic Deﬁnitions
A combinatorial optimization problem Π is either a minimization problem
or a maximization problem. It consists of three components:
(1) A set D
Π
of instances.
393
394 Approximation Algorithms
(2) For each instance I ∈ D
Π
, there is a ﬁnite set S
Π
(I) of candidate
solutions for I.
(3) Associated with each solution σ ∈ S
Π
(I) to an instance I in D
Π
,
there is a value f
Π
(σ) called the solution value for σ.
If Π is a minimization problem, then an optimal solution σ
∗
for an in-
stance I ∈ D
Π
has the property that for all σ ∈ S
Π
(I), f
Π
(σ
∗
) ≤ f
Π
(σ). An
optimal solution for a maximization problem is deﬁned similarly. Through-
out this chapter, we will denote by OPT(I) the value f
Π
(σ
∗
).
An approximation algorithm A for an optimization problem Π is a (poly-
nomial time) algorithm such that given an instance I ∈ D
Π
, it outputs some
solution σ ∈ S
Π
(I). We will denote by A(I) the value f
Π
(σ).
Example 15.1 In this example, we illustrate the above deﬁnitions. Consider
the problem bin packing: Given a collection of items of sizes between 0 and 1, it
is required to pack these items into the minimum number of bins of unit capacity.
Obviously, this is a minimization problem. The set of instances D
Π
consists of
all sets I = ¦s
1
, s
2
, . . . , s
n
¦, such that for all j, 1 ≤ j ≤ n, s
j
is between 0 and 1.
The set of solutions S
Π
consists of a set of subsets σ = ¦B
1
, B
2
, . . . , B
k
¦ which is
a disjoint partition of I such that for all j, 1 ≤ j ≤ k,
¸
s∈B
j
s ≤ 1.
Given a solution σ, its value f(σ) is simply [σ[ = k. An optimal solution for
this problem is that solution σ having the least cardinality. Let A be (the triv-
ial) algorithm that assigns one bin for each item. Then, by deﬁnition, A is an
approximation algorithm. Clearly, this is not a good approximation algorithm.
Throughout this chapter, we will be interested in optimization problems
as opposed to decision problems. For example, the decision problem version
of the bin packing problem has also as input a bound K, and the solution
is either yes if all items can be packed using at most K bins, and no
otherwise. Clearly, if a decision problem is NP-hard, then the optimization
version of that problem is also NP-hard.
15.3 Diﬀerence Bounds
Perhaps, the most we can hope from an approximation algorithm is that
the diﬀerence between the value of the optimal solution and the value of
Diﬀerence Bounds 395
the solution obtained by the approximation algorithm is always constant.
In other words, for all instances I of the problem, the most desirable so-
lution that can be obtained by an approximation algorithm A is such that
[A(I) −OPT(I)[ ≤ K, for some constant K. There are very few NP-hard
optimization problems for which approximation algorithms with diﬀerence
bounds are known. One of them is the following problem.
15.3.1 Planar graph coloring
Let G = (V, E) be a planar graph. By the Four Color Theorem, every planar
graph is four-colorable. It is fairly easy to determine whether a graph is 2-
colorable or not (Exercise 10.3). On the other hand, to determine whether it
is 3-colorable is NP-complete. Given an instance I of G, an approximation
algorithm A may proceed as follows. Assume G is nontrivial, i.e., it has at
least one edge. Determine if the graph is 2-colorable. If it is, then output 2;
otherwise output 4. If G is 2-colorable, then [A(I) −OPT(I)[ = 0. If it is
not 2-colorable, then [A(I) −OPT(I)[ ≤ 1. This is because in the latter
case, G is either 3-colorable or 4-colorable.
15.3.2 Hardness result: the knapsack problem
The problem knapsack is deﬁned as follows (see Sec. 7.6). Given n
items ¦u
1
, u
2
, . . . , u
n
¦ with integer sizes s
1
, s
2
, . . . , s
n
and integer values
v
1
, v
2
, . . . , v
n
, and a knapsack capacity C that is a positive integer, the
problem is to ﬁll the knapsack with some of these items whose total size
is at most C and whose total value is maximum. In other words, ﬁnd a
subset S ⊆ U such that
¸
u
j
∈S
s
j
≤ C and
¸
u
j
∈S
v
j
is maximum.
We will show that there is no approximation algorithm with diﬀerence
bound that solves the knapsack problem. Suppose there is an approxi-
mation algorithm A to solve the knapsack problem with diﬀerence bound
K, i.e., for all instances I of the problem, [A(I) −OPT(I)[ ≤ K, where K
is a positive integer. Given an instance I, we can use algorithm A to output
an optimal solution as follows. Construct a new instance I

such that for
all j, 1 ≤ j ≤ n, s

j
= s
j
and v

j
= (K + 1)v
j
. It is easy to see that any
solution to I

is a solution to I and vice versa. The only diﬀerence is that
396 Approximation Algorithms
the value of the solution for I

is (K+1) times the value of the solution for
I. Since A(I

) = (K + 1)A(I), A(I

) −OPT(I

) ≤ K implies
[A(I) −OPT(I)[ ≤
¸
K
K + 1
¸
= 0.
This means that A always gives the optimal solution, i.e., it solves the knap-
sack problem. Since the knapsack problem is known to be NP-hard, it is
highly unlikely that the approximation algorithm A exists unless NP = P.
(Recall that, by deﬁnition, an approximation algorithm runs in polynomial
time).
15.4 Relative Performance Bounds
Clearly, a diﬀerence bound is the best bound guaranteed by an approxima-
tion algorithm. However, it turns out that very few hard problems possess
such a bound, as exempliﬁed by the knapsack problem for which we have
shown that the problem of ﬁnding an approximation algorithm with a dif-
ference bound is impossible unless NP = P. In this section, we will discuss
another performance guarantee, namely the relative performance guaran-
tee.
Let Π be a minimization problem and I an instance of Π. Let A be
an approximation algorithm to solve Π. We deﬁne the approximation ratio
R
A
(I) to be
R
A
(I) =
A(I)
OPT(I)
.
If Π is a maximization problem, then we deﬁne R
A
(I) to be
R
A
(I) =
OPT(I)
A(I)
.
Thus the approximation ratio is always greater than or equal to one. This
has been done so that we will have a uniform measure for the quality of the
solution produced by A.
The absolute performance ratio R
A
for the approximation algorithm A
is deﬁned by
R
A
= inf¦r [ R
A
(I) ≤ r for all instances I ∈ D
Π
¦.
Relative Performance Bounds 397
The asymptotic performance ratio R
∞
A
for the approximation algorithm A
is deﬁned by
R
∞
A
= inf

r ≥ 1

for some integer N, R
A
(I) ≤ r for all
instances I ∈ D
Π
with OPT(I) ≥ N

.
It turns out that quite a few problems possess approximation algorithms
with relative performance ratios. For some problems, the asymptotic ratio
is more appropriate than the absolute performance ratio. For some others,
both ratios are identical. In the following sections, we will consider some
problems for which an approximation algorithm with constant relative ratio
exists.
15.4.1 The bin packing problem
The optimization version of the bin packing problem can be stated as
follows. Given a collection of items u
1
, u
2
, . . . , u
n
of sizes s
1
, s
2
, . . . , s
n
,
where each s
j
is between 0 and 1, we are required to pack these items into
the minimum number of bins of unit capacity. We list here four heuristics
for the bin packing problem.
• First Fit (FF). In this method, the bins are indexed as 1,2, . . . .
All bins are initially empty. The items are considered for packing
in the order u
1
, u
2
, . . . , u
n
. To pack item u
i
, ﬁnd the least index
j such that bin j contains at most 1 − s
i
, and add item u
i
to the
items packed in bin j.
• Best Fit (BF). This method is the same as the FF method except
that when item u
i
is to be packed, we look for that bin, which is
ﬁlled to level l ≤ 1 −s
i
and l is as large as possible.
• First Fit Decreasing (FFD). In this method, the items are ﬁrst
ordered by decreasing order of size, and then packed using the FF
method.
• Best Fit Decreasing (BFD). In this method, the items are ﬁrst
ordered by decreasing order of size, and then packed using the BF
method.
It is easy to prove that R
FF
< 2, where R
FF
is the absolute performance
ratio of the FF heuristic. Let FF(I) denote the number of bins used by the
FF heuristic to pack the items in instance I, and let OPT(I) be the number
398 Approximation Algorithms
of bins in an optimal packing. First, we note that if FF(I) > 1, then
FF(I) <
¸
2
n
¸
i=1
s
i
¸
. (15.1)
To see this, note that no two bins can be half empty. Suppose for the sake
of contradiction that there are two bins B
i
and B
j
that are half empty,
where i < j. Then, the ﬁrst item u
k
put into bin B
j
is of size 0.5 or less.
But this means that the FF algorithm would have had put u
k
in B
i
instead
of starting a new bin. To see that this bound is achievable, consider the
case when for all i, 1 ≤ i ≤ n, s
i
= 0.5 + , where < 1/(2n) is arbitrarily
small. Then, in this case, the number of bins needed is exactly n, which is
less than n + 2n| = n + 1.
On the other hand, it is easy to see that the minimum number of bins
required in an optimal packing is at least the sum of the sizes of all items.
That is,
OPT(I) ≥
¸
n
¸
i=1
s
i
¸
. (15.2)
Dividing inequality 15.1 by inequality 15.2, we have that
R
FF
(I) =
FF(I)
OPT(I)
< 2.
In the bin packing problem, it is more appropriate to use the asymp-
totic performance ratio, as it is more indicative of the performance of the
algorithm for large values of n. A better bound for the FF heuristic is given
by the following theorem whose proof is lengthy and complex.
Theorem 15.1 For all instances I of the bin packing problem,
FF(I) ≤
17
10
OPT(I) + 2.
It can be shown that the BF heuristic also has a performance ratio of
17/10. The FFD algorithm has a better performance ratio, which is given
by the following theorem:
Relative Performance Bounds 399
Theorem 15.2 For all instances I of the bin packing problem,
FFD(I) ≤
11
9
OPT(I) + 4.
Again, it can be shown that the BFD heuristic also has a performance
ratio of 11/9.
15.4.2 The Euclidean traveling salesman problem
In this section we consider the following problem. Given a set S of n points
in the plane, ﬁnd a tour τ on these points of shortest length. Here, a tour
is a circular path that visits every point exactly once. This problem is a
special case of the traveling salesman problem, and is commonly referred
to as the euclidean minimum spanning tree (emst), which is known to
be NP-hard.
Let p
1
be an arbitrary starting point. An intuitive method would pro-
ceed in a greedy manner, visiting ﬁrst that point closest to p
1
, say p
2
, and
then that point which is closest to p
2
, and so on. This method is referred
to as the nearest neighbor (NN) heuristic, and it can be shown that it does
not result in a bounded performance ratio, i.e., R
NN
= ∞. Indeed, it can
be shown that this method results in the performance ratio
R
NN
(I) =
NN(I)
OPT(I)
= O(log n).
An alternative approximation algorithm satisfying R
A
= 2 can be sum-
marized as follows. First, a minimum cost spanning tree T is constructed.
Next, a multigraph T

is constructed from T by making two copies of each
edge in T. Next, an Eulerian tour τ
e
is found (an Eulerian tour is a cycle
that visits every edge exactly once). Once τ
e
is found, it can easily be
converted into the desired Hamiltonian tour τ by tracing the Eulerian tour
τ
e
and deleting those vertices that have already been visited. Figure 15.1
illustrates the method. The input graph in Fig. 15.1(a) is converted into
an Eulerian multigraph in Fig. 15.1(b). Figure 15.1(c) shows the resulting
tour after bypassing those points that have already been visited.
Call this method the MST (minimum spanning tree) heuristic. We
now show that R
MST
< 2. Let τ
∗
denote an optimal tour. Then, the
length of the constructed minimum spanning tree T is strictly less than the
400 Approximation Algorithms
(a) (b) (c)
start
Fig. 15.1 An illustration of the approximation algorithm for the euclidean minimum
spanning tree.
length of τ
∗
. This is because deleting an edge from τ
∗
results in a spanning
tree. Thus, the length of T

is strictly less than twice the length of τ
∗
.
By the triangle inequality, bypassing those vertices that have already been
visited in τ
e
does not increase the length of the tour (recall that the triangle
inequality states that the sum of the lengths of any two sides in a triangle
is greater than or equal to the length of the third side). It follows that the
length of τ is strictly less than twice the length of τ
∗
. This establishes the
bound R
MST
< 2.
The idea behind the MST approximation algorithm can be improved to
obtain a better performance ratio for this problem. To make T Eulerian,
we do not double its edges. Instead, we ﬁrst identify the set X of vertices
of odd degree. The cardinality of X is always even (Exercise 15.5). Next,
we ﬁnd a minimum weight matching M on the members of X. Finally, we
set T

= T ∪M. Clearly, each vertex in T

has an even degree, and thus T

is Eulerian. Continuing as before, we proceed to ﬁnd τ. Let us refer to this
method as the minimum matching (MM) heuristic. It is described more
precisely in Algorithm etspapprox.
Now we show that the performance ratio of this algorithm is 3/2. Let
τ
∗
be an optimal tour. First, observe that length(T) < length(τ
∗
). Next,
note that length(M) ≤ (1/2)length(τ
∗
). To see this, Let τ

be τ
∗
with all
vertices not in X and their incident edges removed. Then, τ

, which is a
cycle, consists of two matchings M
1
and M
2
on the set of points in X. Since
M is a minimum weight matching, its total weight is less than or equal to
Relative Performance Bounds 401
Algorithm 15.1 etspapprox
Input: An instance I of euclidean minimum spanning tree
Output: A tour τ for instance I.
1. Find a minimum spanning tree T of S.
2. Identify the set X of odd degree in T.
3. Find a minimum weight matching M on X.
4. Find an Eulerian tour τ
e
in T ∪ M.
5. Traverse τ
e
edge by edge and bypass each previously visited vertex.
Let τ be the resulting tour.
either one of M
1
or M
2
. It follows that
length(τ) ≤ length(τ
e
)
= length(T) + length(M)
< length(τ
∗
) +
1
2
length(τ
∗
)
=
3
2
length(τ
∗
).
Thus, for any instance of euclidean minimum spanning tree,
R
MM
(I) =
MM(I)
OPT(I)
<
3
2
.
Since Algorithm etspapprox involves ﬁnding a minimum weight
matching, its time complexity is O(n
3
). Finally, we remark that the above
two approximation algorithms apply to any instance of the general traveling
salesman problem in which the triangle inequality is respected.
15.4.3 The vertex cover problem
Recall that a vertex cover C in a graph G = (V, E) is a set of vertices such
that each edge in E is incident to at least one vertex in C. We have shown in
Sec. 10.4.2 that the problem of deciding whether a graph contains a vertex
cover of size k, where k is a positive integer, is NP-complete. It follows that
the problem of ﬁnding a vertex cover of minimum size is NP-hard.
Perhaps, the most intuitive heuristic that comes to mind is as follows.
Repeat the following step until E becomes empty. Pick an edge e arbitrarily
and add one of its endpoints, say v, to the vertex cover. Next, delete e and
all other edges incident to v. Surely, this is an approximation algorithm that
402 Approximation Algorithms
outputs a vertex cover. However, it can be shown that the performance ratio
of this algorithm is unbounded. Surprisingly, if when considering an edge
e, we add both of its endpoints to the vertex cover, then the performance
ratio becomes 2. The process of picking an edge, adding its endpoints to
the cover, and deleting all edges incident to these endpoints is equivalent
to ﬁnding a maximal matching in G. Note that this matching need not
be of maximum cardinality. This approximation algorithm is outlined in
Algorithm vcoverapprox.
Algorithm 15.2 vcoverapprox
Input: An undirected graph G = (V, E).
Output: A vertex cover C for G.
1. C←¦¦
2. while E ,= ¦¦
3. Let e = (u, v) be any edge in E.
4. C←C ∪ ¦u, v¦
5. Remove e and all edges incident to u or v from E.
6. end while
Algorithm vcoverapprox clearly outputs a vertex cover. We now
show that R
V C
= 2. It is not hard to see that the edges picked in Step 3 of
the algorithm correspond to a maximal matching M. To cover the edges in
M, we need at least [M[ vertices. This implies that the size of an optimal
vertex cover is at least [M[. However, the size of the cover obtained by the
algorithm is exactly 2[M[. It follows that R
V C
= 2. To see that this ratio
is achievable, consider the graph
G = (¦v
1
, v
2
¦, ¦(v
1
, v
2
)¦).
For this graph, an optimal cover is ¦v
1
¦, while the cover obtained by the
algorithm is ¦v
1
, v
2
¦.
15.4.4 Hardness result: the traveling salesman problem
In the last sections, we have presented approximation algorithms with rea-
sonable performance ratios. It turns out, however, that there are many
problems that do not admit bounded performance ratios. For example, the
problems coloring, clique, independent set and the general trav-
eling salesman problem (see Chapter 10) have no known approximation
Relative Performance Bounds 403
algorithms with bounded ratios. Let G = (V, E) be an undirected graph.
By Lemma 10.3, a subset S ⊆ V is an independent set of vertices if and only
if V − S is a vertex cover. Moreover, it can be shown that if S is of max-
imum cardinality, then V − S is of minimum cardinality (Exercise 15.9).
One may be tempted to conclude from this that an approximation algo-
rithm for vertex cover will help in ﬁnding an approximation algorithm
for independent set. This, however, is not the case. To see why, suppose
that G has a minimum vertex cover of size (n/2) − 1. The approximation
algorithm vcoverapprox above for the vertex cover problem will ﬁnd one
of size at most n−2. But the complement of this cover is an independent set
of size 2, while the size of a maximum independent set is, by Exercise 15.9,
exactly n −((n/2) −1) = (n/2) + 1.
Now we turn our attention to the general traveling salesman problem.
The following theorem shows that it is impossible to ﬁnd an approximation
algorithm with bounded ratio for the traveling salesman problem unless
NP = P.
Theorem 15.3 There is no approximation algorithm A for the problem
traveling salesman with R
A
< ∞ unless NP = P.
Proof. Suppose, to the contrary, that there is an approximation algo-
rithm A for the problem traveling salesman with R
A
≤ K, for some
positive integer K. We will show that this can be used to derive a polyno-
mial time algorithm for the problem Hamiltonian cycle, which is known
to be NP-complete (see Chapter 10). Let G = (V, E) be an undirected
graph with n vertices. We construct an instance I of the traveling sales-
man problem as follows. Let V correspond to the set of cities and deﬁne a
distance function d(u, v) for all pairs of cities u and v by
d(u, v) =

1 if (u, v) ∈ E
Kn if (u, v) ∈ E.
Clearly, if G has a Hamiltonian cycle, then OPT(I) = n; otherwise
OPT(I) > Kn. Therefore, since R
A
≤ K, we will have A(I) ≤ Kn if
and only if G has a Hamiltonian cycle. This implies that there exists a
polynomial time algorithm for the problem Hamiltonian cycle. But
this implies that NP = P, which is highly unlikely. To ﬁnish the proof,
note that the construction of instance I of the traveling salesman problem
can easily be achieved in polynomial time. ⁄
404 Approximation Algorithms
15.5 Polynomial Approximation Schemes
So far we have seen that for some NP-hard problems there exist approxi-
mation algorithms with bounded approximation ratio. On the other hand,
for some problems, it is impossible to devise an approximation algorithm
with bounded ratio unless NP = P. On the other extreme, it turns out
that there are problems for which there exists a series of approximation al-
gorithms whose performance ratio converges to 1. Examples of these prob-
lems include the problems knapsack, subset-sum and multiprocessor
scheduling.
Deﬁnition 15.1 An approximation scheme for an optimization problem
is a family of algorithms ¦A

[ > 0¦ such that R
A

≤ 1 +.
Thus, an approximation scheme can be viewed as an approximation
algorithm A whose input is an instance I of the problem and a bound error
such that R
A
(I, ) ≤ 1 +.
Deﬁnition 15.2 A polynomial approximation scheme (PAS) is an ap-
proximation scheme ¦A

¦, where each algorithm A

runs in time that is
polynomial in the length of the input instance I.
Note that in this deﬁnition, A

may not be polynomial in 1/. In the
next section, we will strengthen the deﬁnition of an approximation scheme
so that the algorithms run in time that is also polynomial in 1/. In this sec-
tion, we will investigate a polynomial approximation scheme for the knap-
sack problem.
15.5.1 The knapsack problem
Let U = ¦u
1
, u
2
, . . . , u
n
¦ be a set of items to be packed in a knapsack of
size C. For 1 ≤ j ≤ n, let s
j
and v
j
be the size and value of the jth item,
respectively. Recall that the objective is to ﬁll the knapsack with some
items in U whose total size is at most C and such that their total value is
maximum (see Sec. 7.6). Assume without loss of generality that the size of
each item is not larger than C.
Consider the greedy algorithm that ﬁrst orders the items by decreasing
value to size ratio (v
j
/s
j
), and then considers the items one by one for
packing. If the current item ﬁts in the available space, then it is included,
Polynomial Approximation Schemes 405
otherwise the next item is considered. The procedure terminates as soon
as all items have been considered, or no more items can be included in the
knapsack. This greedy algorithm does not result in a bounded ratio as is
evident from the following instance. Let U = ¦u
1
, u
2
¦, s
1
= 1, v
1
= 2, s
2
=
v
2
= C > 2. In this case, the algorithm will pack only item u
1
, while in
the optimal packing, item u
2
is selected instead. Since C can be arbitrarily
large, the performance ratio of this greedy algorithm is unbounded.
Surprisingly, a simple modiﬁcation of the above algorithm results in
a performance ratio of 2. The modiﬁcation is to also test the packing
consisting of the item of largest possible size only, and then the better
of the two packings is chosen as the output. Call this approximation al-
gorithm knapsackgreedy. This approximation algorithm is outlined in
Algorithm knapsackgreedy. It will be left as an exercise to show that
R
knapsackgreedy
= 2 (Exercise 15.6).
Algorithm 15.3 knapsackgreedy
Input: 2n + 1 positive integers corresponding to item sizes ¦s
1
, s
2
, . . . , s
n
¦,
item values ¦v
1
, v
2
, . . . , v
n
¦ and the knapsack capacity C.
Output: A subset Z of the items whose total size is at most C.
1. Renumber the items so that v
1
/s
1
≥ v
2
/s
2
≥ . . . ≥ v
n
/s
n
.
2. j ←0; K←0; V ←0; Z←¦¦
3. while j < n and K < C
4. j ←j+1
5. if s
j
≤ C −K then
6. Z←Z ∪ ¦u
j
¦
7. K←K +s
j
8. V ←V +v
j
9. end if
10. end while
11. Let Z

.
Now, we describe a polynomial approximation scheme for the knapsack
problem. The idea is quite simple. Let = 1/k for some positive integer k.
Algorithm A

consists of two steps. The ﬁrst step is to choose a subset of
at most k items and put them in the knapsack. The second step is to run
Algorithm knapsackgreedy on the remaining items in order to complete
the packing. These two steps are repeated
¸
k
j=0

n
j

times, once for each
406 Approximation Algorithms
subset of size j, 0 ≤ j ≤ k. In the following theorem, we bound both the
running time and performance ratio of Algorithm A

, for all k ≥ 1.
Theorem 15.4 Let = 1/k for some k ≥ 1. Then, the running time of
Algorithm A

is O(kn
k+1
), and its performance ratio is 1 +.
Proof. Since
¸
k
j=0

n
j

= O(kn
k
) (see Exercise 15.7), the number of
subsets of size at most k is O(kn
k
). The amount of work done in each iter-
ation is O(n), and hence the time complexity of the algorithm is O(kn
k+1
).
Now we bound the performance ratio of the algorithm. Let I be an instance
of the knapsack problem with items U = ¦u
1
, u
2
, . . . , u
n
¦ and C being the
knapsack capacity. Let X be the set of items corresponding to an optimal
solution. If [X[ ≤ k, then there is nothing to prove, as the algorithm will try
all possible k-subsets. So, suppose that [X[ > k. Let Y = ¦u
1
, u
2
, . . . , u
k
¦
be the set of k items of largest value in X, and let Z = ¦u
k+1
, u
k+2
, . . . , u
r
¦
denote the set of remaining items in X, assuming v
j
/s
j
≥ v
j+1
/s
j+1
for all
j, k + 1 ≤ j ≤ r −1. Since the elements in Y are of largest value, we must
have
v
j
≤
OPT(I)
k + 1
for j = k + 1, k + 2, . . . , r. (15.3)
Consider now the iteration in which the algorithm tries the set Y as the
initial k-subset, and let u
m
be the ﬁrst item of Z not included into the
knapsack by the algorithm. If no such item exists, then the output of the
algorithm is optimal. So, assume that u
m
exists. The optimal solution can
be written as
OPT(I) =
k
¸
j=1
v
j
+
m−1
¸
j=k+1
v
j
+
r
¸
j=m
v
j
(15.4)
Let W denote the set of items packed by the algorithm, but not in
¦u
1
, u
2
, . . . , u
m
¦, that were considered by the algorithm before u
m
. In
other words, if u
j
∈ W, then u
j
/ ∈ ¦u
1
, u
2
, . . . , u
m
¦ and v
j
/s
j
≥ v
m
/s
m
.
Now, A(I) can be written as
A(I) ≥
k
¸
j=1
v
j
+
m−1
¸
j=k+1
v
j
+
¸
j∈W
v
j
(15.5)
Fully Polynomial Approximation Schemes 407
Let
C

< 1.
Consequently,
R
k
=
OPT(I)
A(I)
< 1 +
1
k
= 1 +.
⁄
15.6 Fully Polynomial Approximation Schemes
The polynomial approximation scheme described in Sec. 15.5 runs in time
that is exponential in 1/, the reciprocal of the desired error bound. In this
section, we demonstrate an approximation scheme in which the approxi-
mation algorithm runs in time that is also polynomial in 1/. This can
408 Approximation Algorithms
be achieved for some NP-hard problems using a constrained approximation
scheme which we deﬁne below.
Deﬁnition 15.3 A fully polynomial approximation scheme (FPAS) is an
approximation scheme ¦A

¦, where each algorithm A

runs in time that is
polynomial in both the length of the input instance and 1/.
Deﬁnition 15.4 A pseudopolynomial time algorithm is an algorithm that
runs in time that is polynomial in the value of L, where L is the largest
number in the input instance.
Notice that if an algorithm runs in time that is polynomial in log L, then
it is a polynomial time algorithm. Here, log L is commonly referred to as
the size of L. In Chapter 7, we have seen an example of a pseudopolynomial
time algorithm, namely the algorithm for the knapsack problem. The idea
behind ﬁnding an FPAS for an NP-hard problem is typical to all problems
for which a pseudopolynomial time algorithm exists. Starting from such
an algorithm A, scaling and rounding are applied to the input values in an
instance I to obtain an instance I

. Then, the same algorithm A is applied
to the modiﬁed instance I

to obtain an answer that is an approximation
of the optimal solution. In this section, we will investigate an FPAS for the
subset-sum problem.
15.6.1 The subset-sum problem
The subset-sum problem is a special case of the the knapsack problem in
which the item values are identical to their sizes. Thus, the subset-sum
problem can be deﬁned as follows. Given n items of sizes s
1
, s
2
, . . . , s
n
, and
a positive integer C, the knapsack capacity, the objective is to ﬁnd a subset
of the items that maximizes the total sum of their sizes without exceeding
the knapsack capacity C. Incidentally, this problem is a variant of the
partition problem (see Sec. 10.4.3). The algorithm to solve this problem is
almost identical to that for the knapsack problem described in Sec. 7.6. It
is shown below as Algorithm subsetsum.
Clearly, the time complexity of Algorithm subsetsum is exactly the
size of the table, Θ(nC), as ﬁlling each entry requires Θ(1) time. Now, we
develop an approximation algorithm A

. The running
time is now reduced to Θ(nC/K) = Θ(kn
2
). Now, we estimate the error
in the approximate solution. Since an optimal solution cannot contain
more than all the n items, we have the following relationship between the
two optimum values OPT(I) and OPT(I

) corresponding to the original
instance I and the new instance I

.
OPT(I) −K OPT(I

) ≤ Kn.
That is, if we let the approximate solution be K times the output of the
410 Approximation Algorithms
algorithm when presented with instance I

, then we have
OPT(I) −A

(I) ≤ Kn,
or
A

(I) ≥ OPT(I) −Kn = OPT(I) −
C
2(k + 1)
.
We may assume without loss of generality that OPT(I) ≥ C/2. This is
because it is easy to obtain the optimal solution if OPT(I) < C/2 (see
Exercise 15.27). Consequently,
R
A

(I) =
OPT(I)
A

(I)
≤
A

(I) +C/2(k + 1)
A

(I)
≤ 1 +
C/2(k + 1)
OPT(I) −C/2(k + 1)
≤ 1 +
C/2(k + 1)
C/2 −C/2(k + 1)
= 1 +
1
k + 1 −1
= 1 +
1
k
.
Thus, the algorithm’s performance ratio is 1 + , and its running time
is Θ(n
2
/). For example, if we let = 0.1, then we obtain a quadratic
algorithm with performance ratio of 11/10. If we let = 1/n
r
for some
r ≥ 1, then we have an approximation algorithm that runs in time Θ(n
r+2
)
with performance ratio of 1 + 1/n
r
.
15.7 Exercises
15.1. Give an instance I of the bin packing problem such that FF(I) ≥
3
2
OPT(I).
15.2. Give an instance I of the bin packing problem such that FF(I) ≥
5
3
OPT(I).
15.3. Show that the performance ratio of the MST heuristic is achievable.
In other words, give an instance of the Euclidean traveling salesman
problem on which the MST heuristic results in a performance ratio of 2.
Exercises 411
15.4. Show that the performance ratio of the NN approximation algorithm for
the Euclidean traveling salesman problem is unbounded.
15.5. Show that the number of vertices of odd degree in an graph is even.
15.6. Show that the performance ratio of Algorithm knapsackgreedy for the
knapsack problem is 2.
15.7. Show that
¸
k
j=0

n
j

= O(kn
k
).
15.8. Theorem 15.4 states that the running time of Algorithm A

is O(kn
k+1
),
where k = 1/ is part of the input. Explain why this is an exponential
algorithm.
15.9. Let G = (V, E) be an undirected graph. By Lemma 10.3, a subset S ⊆ V
is an independent set of vertices if and only if V − S is a vertex cover
for G. Show that if S is of maximum cardinality, then V −S is a vertex
cover of minimum cardinality.
15.10. Consider the following algorithm for ﬁnding a vertex cover in an undi-
rected graph. First, order the vertices by decreasing order of degree.
Next execute the following step until all edges are covered. Pick a vertex
of highest degree that is incident to at least one edge in the remaining
graph, add it to the cover, and delete all edges incident to that vertex.
Show that this greedy approach does not always result in a vertex cover
of minimum size.
15.11. Show that the performance ratio of the approximation algorithm in Ex-
ercise 15.10 for the vertex cover problem is unbounded.
15.12. Consider the following approximation algorithm for the problem of ﬁnd-
ing a maximum clique in a given graph G. Repeat the following step
until the resulting graph is a clique. Delete from G a vertex that is not
connected to every other vertex in G. Show that this greedy approach
does not always result in a clique of maximum size.
15.13. Show that the performance ratio of the approximation algorithm in Ex-
ercise 15.12 for the maximum clique problem is unbounded.
15.14. Consider the following approximation algorithm for the problem of ﬁnd-
ing a maximum clique in a given graph G. Set C = ¦¦ and repeat the
following step until G has no vertex that is not in C or connected to
every other vertex in C. Add to C a vertex that is not in C and is
connected to every other vertex in C.
15.15. Show that the performance ratio of the heuristic algorithm in Exer-
cise 15.14 for the maximum clique problem is unbounded.
15.16. Give an approximation algorithm for the coloring problem: Find the
minimum number of colors needed to color an undirected graph so that
412 Approximation Algorithms
adjacent vertices are assigned diﬀerent colors. Prove or disprove that its
performance ratio is bounded.
15.17. Give an approximation algorithm for the independent set problem:
Find the maximum number of vertices that are mutually disconnected
from each other. Prove or disprove that its performance ratio is bounded.
15.18. Show that Algorithm vcoverapprox does not always give an optimal
vertex cover by giving a counterexample of a graph consisting of at least
three vertices.
15.19. Give an O(n) time algorithm that ﬁnds a vertex cover in a tree in linear
time.
15.20. Show in more detail that the running time of the polynomial approxi-
mation scheme for the knapsack problem discussed in the proof of The-
orem 15.4 is O(kn
k+1
). You should take into account the time needed
to generate the subsets.
15.21. Consider the optimization version of the set cover problem deﬁned
in Sec. 10.4.3: Given a set X of n elements, a family T of subsets of
X, ﬁnd a subset ( ⊆ T of minimum size that covers all the elements
in X. An approximation algorithm to solve this problem is outlined as
follows. Initialize S = X, and C = ¦¦, and repeat the following step
until S = ¦¦. Choose a subset Y ∈ T that maximizes [Y ∩ S[, add Y to
( and set S = S −Y . Show that this greedy algorithm does not always
produce a set cover of minimum size.
15.22. Show that the performance ratio of the approximation algorithm de-
scribed in Exercise 15.21 for the set cover problem is unbounded.
15.23. Show that the performance ratio of the approximation algorithm de-
scribed in Exercise 15.21 for the set cover problem is O(log n).
15.24. Consider the optimization version of the multiprocessor scheduling
problem deﬁned in Sec. 10.4.3: Given n jobs J
1
, J
2
, . . . , J
n
, each having
a run time t
i
and a positive integer m (number of processors), schedule
those jobs on the m processors so as to minimize the ﬁnishing time. The
ﬁnishing time is deﬁned to be the maximum execution time among all
the m processors. An approximation algorithm to solve this problem
is similar to the FF algorithm: The jobs are considered in their order
J
1
, J
2
, . . . , J
n
, each job is assigned to the next available processor (ties
are broken arbitrarily). In other words, the next job is assigned to that
processor with the least ﬁnishing time. Show that the performance ratio
of this algorithm is 2 −1/m.
15.25. Show that the 2 − 1/m bound of the approximation algorithm in Exer-
cise 15.24 is tight by exhibiting an instance that achieves this ratio.
15.26. Consider modifying the approximation algorithm described in Exer-
cise 15.24 for the multiprocessor scheduling problem by ﬁrst or-
Bibliographic notes 413
dering the jobs by decreasing value of their run times. Prove that in this
case the performance ratio becomes
4
3
−
1
3m
.
15.27. Consider the subset-sum problem discussed in Sec. 15.6.1. Show that if
OPT(I) < C/2, then it is straighforward to obtain the optimal solution.
(Hint: Show that
¸
n
j=1
s
j
< C).
15.8 Bibliographic notes
Garey and Johnson (1979) and Motwani (1992) provide a good exposition
to approximation algorithms. Other introductions to approximation algo-
rithms can also be found in Horowitz and Sahni (1978) and Papadimitriou
and Steiglitz (1982). The 17/10 bound for the FF algorithm for the bin
packing problem is due to Johnson et al. (1974). The 11/9 bound for the
FFD algorithm for the bin packing problem is due to Johnson (1973). The
approximation algorithm for the traveling salesman problem appears
in Rosenkrantz et al. (1977). The 3/2 bound is due to Christoﬁdes (1976).
Lawler et al.. (1985) provide an extensive treatment of the traveling
salesman problem. The PAS for the knapsack problem can be found in
Sahni (1975). The FPAS for the subset-sum problem is based on that for
the knapsack problem due to Ibarra and Kim (1975). Sahni (1977) gives
general techniques for constructing PAS and FPAS. An asymptotic PAS for
the bin packing problem is given by Vega and Lueker (1981).
414
PART 6
Iterative Improvement for
Domain-Speciﬁc Problems
415
416
417
In this part of the book, we study an algorithm design technique that
we will refer to as iterative improvement. In its simplest form, this tech-
nique starts with a simple-minded (usually a greedy) solution and continues
to improve on that solution in stages until an optimal solution is found.
One more aspect of problem-speciﬁcity characterizes this technique. Some
marking characteristics of the iterative improvement technique are in order.
First, devicing new data structures to meet the data access requirements
of the algorithm eﬀectively, e.g. splay trees and Fibonacci heaps. Second,
the introduction of innovative analysis techniques to carefully account for
the true cost of the computation. This will be evident when, for exam-
ple, counting the number of phases or augmentations in network ﬂow and
matching algorithms. Third, exploiting the problem speciﬁc observations
to improve upon the existing solution.
As examples of this design technique, we will study in detail two prob-
lems: ﬁnding a maximum ﬂow in a network, and ﬁnding a maximum match-
ing in undirected graphs. Both of these problems have received a great
amount of attention by researchers, and as a result many algorithms have
been developed. Beside being interesting in their own right, these problems
arise as subproblems in many practical applications.
For the maximum ﬂow problem, which is the subject of Chapter 16,
we present a sequence of increasingly eﬃcient algorithms, starting from an
algorithm with unbounded time complexity to an algorithm that runs in
cubic time.
Chapter 17 is devoted to the problem of ﬁnding a maximum matching
in an undirected graph. We will give algorithms for bipartite graphs and
general graphs. We close this chapter with an elegant matching algorithm
in bipartite graphs that runs in time O(n
2.5
).
418
Chapter 16
Network Flow
16.1 Introduction
Let G = (V, E) be a directed graph with two distinguished vertices s
and t called, respectively, the source and sink, and a capacity function
c(u, v) deﬁned on all pairs of vertices. Throughout this chapter, the 4-tuple
(G, s, t, c), or simply G, will denote a network. Also, n and m will denote,
respectively, the number of vertices and edges in G, that is, n = [V [ and
m = [E[. In this chapter, we consider the problem of ﬁnding a maximum
ﬂow in a given network (G, s, t, c) from s to t. This problem is called the
max-ﬂow problem. We will present a series of algorithms starting from a
method of unbounded time complexity to an algorithm that runs in time
O(n
3
).
16.2 Preliminaries
Let G = (V, E) be a directed graph with two distinguished vertices s and
t called, respectively, the source and sink, and a capacity function c(u, v)
deﬁned on all pairs of vertices with c(u, v) > 0 if (u, v) ∈ E and c(u, v) = 0
otherwise.
Deﬁnition 16.1 A ﬂow in G is a real-valued function f on vertex pairs
having the following three conditions:
C1. Skew symmetry. ∀ u, v ∈ V, f(u, v) = −f(v, u). We say there is a ﬂow
from u to v if f(u, v) > 0.
419
420 Network Flow
C2. Capacity constraints. ∀ u, v ∈ V, f(u, v) ≤ c(u, v). We say edge (u, v)
is saturated if f(u, v) = c(u, v).
C3. Flow conservation. ∀ u ∈ V −¦s, t¦,
¸
v∈V
f(u, v) = 0. In other words,
the net ﬂow (total ﬂow out minus total ﬂow in) at any interior vertex is 0.
C4. ∀ v ∈ V, f(v, v) = 0.
Deﬁnition 16.2 A cut ¦S, T¦ is a partition of the vertex set V into two
subsets S and T such that s ∈ S and t ∈ T. The capacity of the cut ¦S, T¦,
denoted by c(S, T), is
c(S, T) =
¸
u∈S,v∈T
c(u, v).
The ﬂow across the cut ¦S, T¦, denoted by f(S, T), is
f(S, T) =
¸
u∈S,v∈T
f(u, v).
Thus, the ﬂow across the cut ¦S, T¦ is the sum of the positive ﬂow on
edges from S to T minus the sum of the positive ﬂow on edges from T to S.
For any vertex u and any subset A ⊆ V , let f(u, A) denote f(¦u¦, A), and
f(A, u) denote f(A, ¦u¦). For a capacity function c, c(u, A) and c(A, u) are
deﬁned similarly.
Deﬁnition 16.3 The value of a ﬂow f, denoted by [f[, is deﬁned to be
[f[ = f(s, V ) =
¸
v∈V
f(s, v).
Lemma 16.1 For any cut ¦S, T¦ and a ﬂow f, [f[ = f(S, T).
Proof. By induction on the number of vertices in S. If S = ¦s¦, then it
is true by deﬁnition. Assume it is true for the cut ¦S, T¦. We show that it
also holds for the cut ¦S∪¦w¦, T −¦w¦¦ for w ∈ T −¦t¦. Let S

(u, v).
The following two lemmas, which appear to be intuitive, provide the
basis for the iterative improvement technique in network ﬂow. Their proofs
are left for the exercises.
Lemma 16.2 Let f be a ﬂow in G and f

the ﬂow in the residual graph
R for f. Then the function f +f

is a ﬂow in G of value [f[ +[f

[.
Lemma 16.3 Let f be any ﬂow in G and f
∗
a maximum ﬂow in G. If
R is the residual graph for f, then the value of a maximum ﬂow in R is
[f
∗
[ −[f[.
Deﬁnition 16.5 Given a ﬂow f in G, an augmenting path p is a directed
path from s to t in the residual graph R. The bottleneck capacity of p is
the minimum residual capacity along p. The number of edges in p will be
denoted by [p[.
In Fig. 16.1(b), the path s, a, c, b, d, t is an augmenting path with bot-
tleneck capacity 2. If two additional units of ﬂow are pushed along this
path, then the ﬂow becomes maximum.
Theorem 16.1 (max-ﬂow min-cut theorem). Let (G, s, t, c) be a network
and f a ﬂow in G. The following three statements are equivalent:
(a) There is a cut ¦S, T¦ with c(S, T) = [f[.
(b) f is a maximum ﬂow in G.
(c) There is no augmenting path for f.
Proof. (a)→(b). Since [f[ ≤ c(A, B) for any cut ¦A, B¦, c(S, T) = [f[
implies f is a maximum ﬂow.
(b)→(c). If there is an augmenting path p in G, then [f[ can be increased
by increasing the ﬂow along p, i.e., f is not maximum.
(c)→(a). Suppose there is no augmenting path for f. Let S be the set of
vertices reachable from s by paths in the residual graph R. Let T = V −S.
Then, R contains no edges from S to T. Thus, in G, all edges from S to T
are saturated. It follows that c(S, T) = [f[. ⁄
The proof of the implication (c)→(a) suggests an algorithm for ﬁnding
a minimum cut in a given network.
The Ford-Fulkerson Method 423
16.3 The Ford-Fulkerson Method
Theorem 16.1 suggests a way to construct a maximum ﬂow by iterative im-
provement: One keeps ﬁnding an augmenting path arbitrarily and increases
the ﬂow by its bottleneck capacity. This is known as the Ford-Fulkerson
method.
Algorithm 16.1 ford-fulkerson
Input: A network (G, s, t, c).
Output: A ﬂow in G.
1. Initialize the residual graph: Set R = G.
2. for each edge (u, v) ∈ E
3. f(u, v)←0
4. end for
5. while there is an augmenting path p = s, . . . , t in R
6. Let ∆ be the bottleneck capacity of p.
7. for each edge (u, v) in p
8. f(u, v)←f(u, v) + ∆
9. end for
10. Update the residual graph R.
11. end while
Step 1 initializes the residual graph to the original network. Step 2 ini-
tializes the ﬂow in G to the zero ﬂow. The while loop is executed for each
augmenting path found in the residual graph R. Each time an augment-
ing path is found, its bottleneck capacity ∆ is computed and the ﬂow is
increased by ∆. This is followed by updating the residual graph R. Updat-
ing R may result in the addition of new edges or the deletion of some of the
existing ones. It should be emphasized that the selection of the augmenting
path in this method is arbitrary.
The Ford-Fulkerson method may not halt if the capacities are irrational.
If the ﬂow does converge, however, it may converge to a value that is not
necessarily maximum. If the capacities are integers, this method always
computes the maximum ﬂow f
∗
in at most [f
∗
[ steps, since each augmenta-
tion increases the ﬂow by at least 1. As each augmenting path can be found
in O(m) time (e.g. using depth-ﬁrst-search), the overall time complexity of
this method (when the input capacities are integers) is O(m[f
∗
[). Notice
that this time complexity is dependent on the input values. As an example,
consider the network shown in Fig. 16.2(a). If the method alternately se-
424 Network Flow
lects the augmenting paths s, a, b, t and s, b, a, t, the number of augmenting
steps is 1000. The ﬁrst two residual graphs are shown in Figs. 16.2(b) and
(c).
/
t
·
o
(c)
499
499
1
1
1
499
499
1
1
/
t
·
o
1
(b)
1
1
499 500
500
499
t
·
o
/
(a)
500
500
500
500
1
Fig. 16.2 An example of a graph on which the ford-fulkerson method performs badly.
16.4 Maximum Capacity Augmentation
In this section we consider improving the Ford-Fulkerson method by select-
ing among all possible augmenting paths that path with maximum bot-
tleneck capacity. This heuristic is due to Edmonds and Karp. As an ex-
ample, consider the original graph with zero ﬂow of the network shown in
Fig. 16.1(a). According to this heuristic, the augmenting path s, a, c, b, d, t
with bottleneck capacity 6 is ﬁrst selected. This is followed by choosing
the augmenting path s, b, c, t with bottleneck capacity 2. If the augmenting
path s, b, c, d, t with bottleneck capacity 2 is next selected, then the ﬂow
becomes maximum. As to the network shown in Fig. 16.2, a maximum ﬂow
can be found using this method after exactly two augmentations.
To analyze the time complexity of this method, which we will refer to
as the maximum capacity augmentation (mca) method, we ﬁrst show that
there always exists a sequence of at most m augmentations that lead to a
maximum ﬂow. Next, we show that if the input capacities are integers, then
its time complexity is polynomial in the input size, which is a signiﬁcant
improvement on the Ford-Fulkerson method.
Lemma 16.4 Starting from the zero ﬂow, there is a sequence of at most
m augmentations that lead to a maximum ﬂow.
Proof. Let f
∗
be a maximum ﬂow. Let G
∗
be the subgraph of G induced
by the edges (u, v) such that f
∗
(u, v) > 0. Initialize i to 1. Find a path p
i
Maximum Capacity Augmentation 425
from s to t in G
∗
. Let ∆
i
be the bottleneck capacity of p
i
. For every edge
(u, v) on p
i
, reduce f
∗
(u, v) by ∆
i
deleting those edges whose ﬂow becomes
zero. Increase i by 1 and repeat the above procedure until t is no longer
reachable from s. This procedure halts after at most m steps, since at least
one edge is deleted in each iteration. It produces a sequence of augmenting
paths p
1
, p
2
, . . . with ﬂows ∆
1
, ∆
2
, . . .. Now, beginning with the zero ﬂow,
push ∆
1
units along p
1
, ∆
2
units along p
2
, . . . to construct a maximum ﬂow
in at most m steps. ⁄
This lemma is not constructive in the sense that it does not provide
a way of ﬁnding this sequence of augmenting paths; it only proves the
existence of such a sequence.
Theorem 16.2 If the edge capacities are integers, then the mca con-
structs a maximum ﬂow in O(mlog c
∗
) augmenting steps, where c
∗
is the
maximum edge capacity.
Proof. Let R be the residual graph corresponding to the initial zero ﬂow.
Since the capacities are integers, there is a maximum ﬂow f
∗
that is an in-
teger. By Lemma 16.4, f
∗
can be achieved in at most m augmenting paths,
and hence there is an augmenting path p in R with bottleneck capacity at
least f
∗
/m. Consider a sequence of 2m consecutive augmentations using
the mca heuristic. One of these augmenting paths must have bottleneck
capacity of f
∗
/2m or less. Thus, after at most 2m augmentations, the
maximum bottleneck capacity is reduced by a factor of at least 2. After at
most 2m more augmentations, the maximum bottleneck capacity is reduced
further by a factor of at least 2. In general, after at most 2km augmenta-
tions, the maximum bottleneck capacity is reduced by a factor of at least
2
k
. Since the maximum bottleneck capacity is at least one, k cannot exceed
log c
∗
. This means the number of augmentations is O(mlog c
∗
). ⁄
A path of maximum bottleneck capacity can be found in O(n
2
) time us-
ing a modiﬁcation of Dijkstra’s algorithm for the single-source shortest path
problem (See ection 8.2). Therefore, the mca heuristic ﬁnds a maximum
ﬂow in O(mn
2
log c
∗
) time.
The time complexity is now polynomial in the input size. However, it is
undesirable that the running time of an algorithm be dependent on its input
values. This dependence will be removed using the algorithms presented in
the following three sections.
426 Network Flow
16.5 Shortest Path Augmentation
In this section we consider another heuristic, also due to Edmonds and
Karp, that puts some order on the selection of augmenting paths. It results
in a time complexity that is not only polynomial, but also independent of
the input values.
Deﬁnition 16.6 The level of a vertex v, denoted by level(v), is the least
number of edges in a path from s to v. Given a directed graph G = (V, E),
the level graph L is (V, E

), where E

= ¦(u, v) [ level(v) = level(u) + 1¦.
Given a directed graph G and a source vertex s, its level graph L can
easily be constructed using breadth-ﬁrst search. As an example of the
construction of the level graph, see Fig. 16.3(a) and (b). In this ﬁgure, the
graph shown in (b) is the level graph of the network shown in (a). Here,
¦s¦, ¦a, b¦, ¦c, d¦, ¦t¦ constitute levels 0, 1, 2 and 3, respectively. Observe
that edges (a, b), (b, a) and (d, c) are not present in the level graph, as they
connect vertices in the same level. Also the edge (c, b) is not included since
it is directed from a vertex of higher level to a vertex of lower level.
This heuristic, which we will refer to as minimum path length augmen-
tation (mpla) method, selects an augmenting path of minimum length and
increases the current ﬂow by an amount equal to the bottleneck capacity
of that path. The algorithm starts by initializing the ﬂow to the zero ﬂow
and setting the residual graph R to the original network. It then proceeds
in phases. Each phase consists of the following two steps.
(1) Compute the level graph L from the residual graph R. If t is not
in L, then halt; otherwise continue.
(2) As long as there is a path p from s to t in L, augment the current
ﬂow by p, remove saturated edges from L and R and update them
accordingly.
Note that augmenting paths in the same level graph are of the same
length. Moreover, as will be shown later, the length of an augmenting
path in any phase after the ﬁrst is strictly longer than the length of an
augmenting path in the preceding phase. The algorithm terminates as soon
as t does not appear in the newly constructed level graph. An outline of
the algorithm is shown as Algorithm mpla (see Fig. 16.3 for an example).
To analyze the running time of the algorithm, we need the following lemma:
Shortest Path Augmentation 427
Algorithm 16.2 mpla
Input: A network (G, s, t, c).
Output: The maximum ﬂow in G.
1. for each edge (u, v) ∈ E
2. f(u, v)←0
3. end for
4. Initialize the residual graph: Set R = G.
5. Find the level graph L of R.
6. while t is a vertex in L
7. while t is reachable from s in L
8. Let p be a path from s to t in L.
9. Let ∆ be the bottleneck capacity on p.
10. Augment the current ﬂow f by ∆.
11. Update L and R along the path p.
12. end while
13. Use the residual graph R to compute a new level graph L.
14. end while
Lemma 16.5 The number of phases in the mpla algorithm is at most n.
Proof. We show that the number of level graphs computed using the
algorithm is at most n. First, we show that the sequence of lengths of
augmenting paths using the mpla algorithm is strictly increasing. Let p be
any augmenting path in the current level graph. After augmenting using
p, at least one edge will be saturated and will disappear in the residual
graph. At most [p[ new edges will appear in the residual graph, but they
are back edges, and hence will not contribute to a shortest path from s to
t. There can be at most m paths of length [p[ since each time an edge in
the level graph disappears. When t is no longer reachable from s in the
level graph, any augmenting path must use a back edge or a cross edge, and
hence must be of length strictly greater than [p[. Since the length of any
augmenting path is between 1 and n − 1, the number of level graphs used
for augmentations is at most n−1. Since one more level graph is computed
in which t does not appear, the total number of level graphs computed is
at most n. ⁄
The running time of the mpla algorithm is computed as follows. Since
there can be at most m augmentations along paths of the same length,
and since by Lemma 16.5 the number of level graphs computed that are
428 Network Flow
t
c
/ d
·
o
4
19
7
11
12
11
12
Final flow.
(j)
/ d
·
o
3
2
4
Third level graph.
(i)
t
c
/
d
·
o
8/7
7/7
10/7
9/7
4
(g)
t
c
/ d
·
o
4
1
12
9
7 10
2
Residual graph.
12
19
11
11
4
3
4
(h)
t
c
/ d
·
o
4
8
12
9
7 10
9
Residual graph.
12
12
4
4
4
10
4
(e)
t
c
/ d
·
o
8
7
10
9
4
Second level graph.
(f)
t
c
/ d
·
o
4/4
14/4
13/4
Augment ·, /, d, t.
(d)
t
c
/ d
·
o
20/12
12/12
16/12
Augment ·, o, c, t.
(c)
Input graph.
t
/ d
c
·
o
4
20
7
14
9
12
4 10
13
16
(a) (b)
First level graph.
t
c
/ d
·
o
4
20
14
12
13
16
Augment ·, /, d, c, t.
Fig. 16.3 Example of the mpla algorithm.
used for augmenting is at most n − 1, the number of augmenting steps is
at most (n − 1)m. Finding a shortest augmenting path in the level graph
takes O(m) time using breadth-ﬁrst search. Thus, the total time needed
Dinic’s Algorithm 429
to compute all augmenting paths is O(nm
2
). Computing each level graph
takes O(m) using breadth-ﬁrst search, and hence the total time required to
compute all level graphs is O(nm). It follows that the overall running time
of Algorithm mpla is O(nm
2
).
As to the correctness of the algorithm, note that after computing at
most n−1 level graphs, there are no more augmenting paths in the original
network. By Theorem 16.1, this implies that the ﬂow is maximum. Hence,
we have the following theorem:
Theorem 16.3 The mpla algorithm ﬁnds a maximum ﬂow in a network
with n vertices and m edges in O(nm
2
) time.
16.6 Dinic’s Algorithm
In Sec. 16.5, it was shown that ﬁnding the maximum ﬂow can be achieved
in O(nm
2
) time. In this section we show that the time complexity can be
reduced to O(mn
2
) using a method due to Dinic. In the mpla algorithm,
after a level graph is computed, augmenting paths are found individually.
In contrast, the algorithm in this section ﬁnds all these augmenting paths
more eﬃciently, and this is where the improvement in the running time
comes from.
Deﬁnition 16.7 Let G be a network and H a subgraph of G containing
both s and t. A ﬂow f in H is a blocking ﬂow (with respect to H) if every
path in H from s to t contains at least one saturated edge.
In Fig. 16.4(c), the ﬂow is a blocking ﬂow with respect to the level
graph shown in (b). Dinic’s method is shown in Algorithm dinic. As in
the mpla algorithm, Dinic’s algorithm is divided into at most n phases.
Each phase consists of ﬁnding a level graph, a blocking ﬂow with respect
to that level graph and increasing the current ﬂow by that blocking ﬂow.
By Lemma 16.5, the number of phases is at most n. Each iteration of the
outer while loop corresponds to one phase. The intermediate while loop
is essentially a depth-ﬁrst search in which augmenting paths are found and
used to increase the ﬂow. Here, p = s, . . . , u is the current path found so
far. There are two basic operations in the inner while loop. If u, which
is the end of the current path, is not t and there is at least one edge out
430 Network Flow
Algorithm 16.3 dinic
Input: A network (G, s, t, c).
Output: The maximum ﬂow in G.
1. for each edge (u, v) ∈ E
2. f(u, v)←0
3. end for
4. Initialize the residual graph: Set R = G.
5. Find the level graph L of R.
6. while t is a vertex in L
7. u ←s
8. p ←u
9. while outdegree(s) > 0 ¦begin phase¦
10. while u ,= t and outdegree(s) > 0
11. if outdegree(u) > 0 then ¦advance¦
12. Let (u, v) be an edge in L.
13. p←p, v
14. u←v
15. else ¦retreat¦
16. Delete u and all adjacent edges from L.
17. Remove u from the end of p.
18. Set u to the last vertex in p (u may be s).
19. end if
20. end while
21. if u = t then ¦augment¦
22. Let ∆ be the bottleneck capacity along p. Augment
the current ﬂow along p by ∆. Adjust capacities
along p in both residual graph and level graph, delet-
ing saturated edges. Set u to the last vertex on p
reachable from s. Note that u may be s.
23. end if
24. end while
25. Compute a new level graph L from the current residual graph R.
26. end while
of u, say (u, v), then an advance operation takes place. This operation
consists of appending v to p and making it the current endpoint of p. If,
on the other hand, u is not t and there is no edge out of it, a retreat
operation takes place. This operation simply amounts to removing u from
the end of p and removing it and all adjacent edges in the current level
graph L, as there cannot be any augmenting path that passes by u. The
inner while loop terminates if either t is reached or the search backs up to
s and all edges out of s have been explored. If t is reached, then this is an
The MPM Algorithm 431
indication that an augmenting path has been discovered and augmenting
by that path is carried out in the steps following the inner while loop. If,
on the other hand, s has been reached and all edges out of it have been
deleted, then no augmentation takes place and processing the current level
graph is complete. An example of the execution of the algorithm is given
in Fig. 16.4.
We compute the running time in each phase as follows. The number
of augmentations is at most m since at least one edge of the level graph is
deleted in each augmentation. Each augment costs O(n) time to update
the ﬂow values and to delete edges in both the level graph, the residual
graph and the path p used in the algorithm and possibly to add edges to
the residual graph. Hence, the total cost of all augments in each phase is
O(mn). The number of retreats (the else part of the inner while loop) is
at most n −2 since each retreat results in the deletion of one vertex other
than s or t. The total number of edges deleted from the level graph in
the retreats is at most m. This means that the total cost of all retreats
is O(m + n) in each phase. The number of advances (the if part of the
inner while loop) before each augment or retreat cannot exceed n −1; for
otherwise one vertex will be visited more than once before an augment
or retreat. Consequently, the total number of advances is O(mn) in each
phase. It follows that the overall cost of each phase is O(mn), and since
there are at most n phases, the overall running time of the algorithm is
O(mn
2
).
As to the correctness of the algorithm, note that after computing at
most n−1 level graphs, there are no more augmenting paths in the residual
graph. By Theorem 16.1, this implies that the ﬂow is maximum. Hence,
we have the following theorem:
Theorem 16.4 Dinic’s algorithm ﬁnds a maximum ﬂow in a network
with n vertices and m edges in O(mn
2
) time.
16.7 The MPM Algorithm
In this section we outline an O(n
3
) time algorithm to ﬁnd the maximum ﬂow
in a given network. The algorithm is due to Malhotra, Pramodh-Kumar
and Maheshwari. It is an improvement on Dinic’s algorithm. The O(n
3
)
bound is due to a faster O(n
2
) time method for computing a blocking ﬂow.
432 Network Flow
Finally, t is not reachable
from ·. Hence, maximum
flow of 15 is achievd.
Input graph.
t
c
/ d
·
o
4
13
9
7
9
8
1
2 10
9
(a)
4
4
4
t
c
/ d
·
o
7
7
7
(c)
A blocking flow.
t
c
/ d
·
o
4
7
9
1 10
Residual graph.
7
7
4
4
2
4
6
5
2
(d)
2
Second level graph.
d
t
c
b
s
a
5
4
2
6
1
(e)
A blocking flow.
d
t
c
/
·
o
3
2
2
3
1
1
(f)
Residual graph.
t
c
/ d
·
o
7
8
9
1
12
10
7
4
6
2
3
2
1
(g)
Third level graph.
/ d
t
c
·
o
9
2
3
2
1 (h)
t
d
c
/
·
o
1
1
1
1
1 (i)
A blocking flow.
Residual graph.
/ d
7
t
c
·
o
8
9
8 1
12
11
7
4
1
2
1
1
(j)
Final flow.
t
c
/ d
·
o
4
11
9
7
8
7
1
2
1
(k)
(b)
First level graph.
t
c
/ d
·
o
9
7
8
9
4
13
Fig. 16.4 Example of Dinic’s algorithm.
In this section we will consider only the method of ﬁnding such a blocking
ﬂow. The rest of the algorithm is similar to Dinic’s algorithm. For this, we
need the following deﬁnition.
The MPM Algorithm 433
Deﬁnition 16.8 For a vertex v in a network G, we deﬁne the throughput
of v as the minimum of the total capacity of incoming edges and the total
capacity of outgoing edges. That is, for v ∈ V −¦s, t¦,
throughput(v) = min

¸
u∈V
c(u, v),
¸
u∈V
c(v, u)
¸
.
The throughputs of s and t are deﬁned by
throughput(s) =
¸
v∈V −{s}
c(s, v) and throughput(t) =
¸
v∈V −{t}
c(v, t).
As in Dinic’s algorithm, updating the residual graph, computing the
level graph and ﬁnding a blocking ﬂow comprise one phase of the algorithm.
Finding a blocking ﬂow from the level graph L can be described as follows.
First, we ﬁnd a vertex v such that g = throughput(v) is minimum among
all other vertices in L. Next, we “push” g units of ﬂow from v all the way
to t and “pull” g units of ﬂow all the way from s. When pushing a ﬂow
out of a vertex v, we saturate some of its outgoing edges to their capacity
and leave at most one edge partially saturated. We then delete all outgoing
edges that are saturated. Similarly, when pulling a ﬂow into a vertex v,
we saturate some of its incoming edges to their capacity and leave at most
one edge partially saturated. We then delete all incoming edges that are
saturated. Either all incoming edges or all outgoing edges will be saturated.
Consequently, vertex v and all its adjacent edges are removed from the level
graph and the residual graph R is updated accordingly. The ﬂow out of
v is pushed through its outgoing edges to (some of) its adjacent vertices
and so on until t is reached. Note that this is always possible, as v has
minimum throughput among all other vertices in the current level graph.
Similarly, the ﬂow into v is propagated backward until s is reached. Next,
another vertex of minimum throughput is found and the above procedure
is repeated. Since there are n vertices, the above procedure is repeated at
most n −1 times. The method is outlined in Algorithm mpm.
The time required by each phase of the algorithm is computed as follows.
The time required to ﬁnd the level graph L is O(m) using breadth-ﬁrst
search. Finding a vertex of minimum throughput takes O(n) time. Since
this is done at most n − 1 times, the total time required by this step is
O(n
2
). Deleting all saturated edges takes O(m) time. Since at most one
434 Network Flow
Algorithm 16.4 mpm
Input: A network (G, s, t, c).
Output: The maximum ﬂow in G.
1. for each edge (u, v) ∈ E
2. f(u, v)←0
3. end for
4. Initialize the residual graph: Set R = G.
5. Find the level graph L of R.
6. while t is a vertex in L
7. while t is reachable from s in L
8. Find a vertex v of minimum throughput = g.
9. Push g units of ﬂow from v to t.
10. Pull g units of ﬂow from s to v.
11. Update f, L and R.
12. end while
13. Use the residual graph R to compute a new level graph L.
14. end while
edge is partially saturated for each vertex, the time required to partially
saturate edges in each iteration of the inner while loop takes O(n) time.
Since there are at most n − 1 iterations of the inner while loop, the total
time required to partially saturate edges is O(n
2
). It follows that the total
time required to push ﬂow from v to t and to pull ﬂow from s to v is O(n
2
).
The time required to update the ﬂow function f and the residual graph R
is no more than the time required to push and pull ﬂows, i.e., O(n
2
). As a
result, the overall time required by each phase is O(n
2
+m) = O(n
2
).
As there are at most n phases (in the ﬁnal phase, t is not a vertex of
L), the overall time required by the algorithm is O(n
3
). Finally, note that
after computing at most n −1 level graphs, there are no more augmenting
paths in the residual graph. By Theorem 16.1, this implies that the ﬂow is
maximum. Hence, we have the following theorem:
Theorem 16.5 The mpm algorithm ﬁnds a maximum ﬂow in a network
with n vertices and m edges in O(n
3
) time.
16.8 Exercises
16.1. Prove Lemma 16.2.
Exercises 435
16.2. Prove Lemma 16.3.
16.3. Let f be a ﬂow in a network G and f

the ﬂow in the residual graph R
for f. Prove or disprove the following claim. If f

is a maximum ﬂow in
R, then f + f

is a maximum ﬂow in G. The function f + f

is deﬁned
on page 421.
16.4. Prove or disprove the following statement. If all capacities in a net-
work are distinct, then there exists a unique ﬂow function that gives the
maximum ﬂow.
16.5. Prove or disprove the following statement. If all capacities in a network
are distinct, then there exists a unique min-cut that separates the source
from the sink.
16.6. Explain how to solve the max-ﬂow problem with multiple sources and
multiple edges.
16.7. Give an O(m) time algorithm to construct the residual graph of a given
network with positive edge capacities.
16.8. Show how to ﬁnd eﬃciently an augmenting path in a given residual
graph.
16.9. Adapt the ford-fulkerson algorithm to the case where the vertices
have capacities as well.
16.10. Give an eﬃcient algorithm to ﬁnd a path of maximum bottleneck capac-
ity in a given directed acyclic graph.
16.11. Give an eﬃcient algorithm to ﬁnd the level graph of a given directed
acyclic graph.
16.12. Show by example that a blocking ﬂow in the level graph of a residual
graph need not be a blocking ﬂow in the residual graph.
16.13. Let G = (V, E) be a directed acyclic graph, where [V [ = n. Give an
algorithm to ﬁnd a minimum number of directed vertex-disjoint paths
which cover all the vertices, i.e., every vertex is in exactly one path.
There are no restrictions on the lengths of the paths, where they start
and end. To do this, construct a ﬂow network G

= (V

, E

), where
V

= ¦s, t¦ ∪ ¦x
1
, x
2
, . . . , x
n
¦ ∪ ¦y
1
, y
2
, . . . , y
n
¦,
E

= ¦(s, x
i
) [ 1 ≤ i ≤ n¦∪¦(y
i
, t) [ 1 ≤ i ≤ n¦∪¦(x
i
, y
j
) [ (v
i
, v
j
) ∈ E¦.
Let the capacity of all edges be 1. Finally, show that the number of
paths which cover V is [V [ −[f[, where f is the maximum ﬂow in G

.
16.14. Let G = (V, E) be a directed graph with two distinguished vertices s, t ∈
V . Give an eﬃcient algorithm to ﬁnd the maximum number of edge-
disjoint paths from s to t.
436 Network Flow
16.15. Let G = (V, E) be an undirected weighted graph with two distinguished
vertices s, t ∈ V . Give an eﬃcient algorithm to ﬁnd a minimum weight
cut that separates s from t.
16.16. Let G = (X ∪Y, E) be a bipartite graph. An edge cover C for G is a set
of edges in E such that each vertex of G is incident to at least one edge
in C. Give an algorithm to ﬁnd an edge cover for G of minimum size.
16.17. Let G = (X ∪ Y, E) be a bipartite graph. Let C be a minimum edge
cover (see the Exercise 16.16) and I a maximum independent set. Show
that [C[ = [I[.
16.18. The vertex connectivity of a graph G = (V, E) is deﬁned as the minimum
number of vertices whose removal disconnects G. Prove that if G has
vertex connectivity k, then [E[ ≥ k [V [/2.
16.9 Bibliographic notes
Part of the material in this chapter is based on Tarjan (1983). A good
and concise coverage of network ﬂow can also be found in Kozen (1992).
Other references for network ﬂow include Even(1979), Lawler(1976), Pa-
padimitriou and Steiglitz (1982). The Ford-Fulkerson method is due to
Ford and Fulkerson (1956). The two heuristics of augmenting by paths
with maximum bottleneck capacity and augmenting by paths of shortest
lengths are due to Edmonds and Karp (1972). Dinic’s algorithm is due to
Dinic (1970). The O(n
3
) MPM algorithm is due to Malhotra, Pramodh-
Kumar and Maheshwari (1978). The O(n
3
) bound remains the best knwon
for general graphs. In the case of sparse graphs, faster algorithms can be
found in Ahuja, Orlin and Tarjan (1989), Galil (1980), Galil and Tardos
(1988), Goldberg and Tarjan (1988), Sleator (1980) and Tardos (1985).
Chapter 17
Matching
17.1 Introduction
In this chapter, we study in detail another example of a problem whose
exisisting algorithms use the iterative improvement design technique: the
problem of ﬁnding a maximum matching in an undirected graph. In its
most general setting, given an undirected graph G = (V, E), the maximum
matching problem asks for a subset M ⊆ E with the maximum number
of nonoverlapping edges, that is, no two edges in M have a vertex in com-
mon. This problem arises in many applications, particularly in the areas of
communication and scheduling. While the problem is interesting in its own
right, it is indispensable as a building block in the design of more complex
algorithms. That is, the problem of ﬁnding a maximum matching is often
used as a subroutine in the implementation of many practical algorithms.
17.2 Preliminaries
Let G = (V, E) be a connected undirected graph. Throughout this chapter,
we will let n and m denote, respectively, the number of vertices and edges
in G, that is, n = [V [ and m = [E[.
A matching in G is a subset M ⊆ E such that no two edges in M have a
vertex in common. We will assume throughout this chapter that the graph
is connected, and hence the modiﬁer “connected” will be dropped. An edge
e ∈ E is matched if it is in M, and unmatched or free otherwise. A vertex
v ∈ V is matched if it is incident to a matched edge, and unmatched or
free otherwise. The size or of a matching M, i.e., the number of matching
437
438 Matching
edges in it, will be denoted by [M[. A maximum matching in a graph is
a matching of maximum cardinality. A perfect matching is one in which
every vertex in V is matched. Given a matching M in an undirected graph
G = (V, E), an A alternating path p with respect to M is a simple path that
consists of alternating matched and unmatched edges. The length of p is
denoted by [p[. If the two endpoints of an alternating path coincide, then
it is called an alternating cycle. An alternating path with respect to M is
called an augmenting path with respect to M if all the matched edges in p
are in M and its endpoints are free. Clearly, the number of edges in an
augmenting path is odd, and as a result, it cannot be an alternating cycle.
These deﬁnitions are illustrated in Fig. 17.1 in which matched edges are
shown as jagged edges.
/ o c
;
j
d
c o
/
| I ,
Fig. 17.1 A matching in an undirected graph.
In Fig. 17.1, M = ¦(b, c), (f, g), (h, l), (i, j)¦ is a matching. The edge (a, b)
is unmatched or free and the edge (b, c) is matched. Vertex a is free and
vertex b is matched. The path a, b, c, d is an alternating path. It is also
an augmenting path (with respect to M). Another augmenting path with
respect to M is a, b, c, g, f, e. Clearly, the matching M is neither maximum
nor perfect.
Let M
1
and M
2
be two matchings in a graph G. Then
M
1
⊕M
2
= (M
1
∪ M
2
) −(M
1
∩ M
2
)
= (M
1
−M
2
) ∪ (M
2
−M
1
).
That is, M
1
⊕ M
2
is the set of edges that are in M
1
or in M
2
but not in
both. Consider the matching shown in Fig. 17.1 and the augmenting path
p = a, b, c, g, f, e.
Reversing the roles of edges in p (matched to unmatched and vice-versa)
Preliminaries 439
/ o c
;
j
d
c o
/
| I ,
Fig. 17.2 An augmented matching.
results in the matching shown in Fig. 17.2. Moreover, the size of the new
matching is exactly the size of the old matching plus one. This illustrates
the following lemma whose proof is easy:
Lemma 17.1 Let M be a matching and p an augmenting path with
respect to M. Then M ⊕p is a matching of size [M ⊕p[ = [M[ + 1.
The following corollary characterizes a maximum matching.
Corollary 17.1 A matching M in an undirected graph G is maximum
if and only if G contains no augmenting paths with respect to M.
Theorem 17.1 Let M
1
and M
2
be two matchings in an undirected G
such that [M
1
[ = r, [M
2
[ = s and s > r. Then, M
1
⊕M
2
contains at least
k = s −r vertex-disjoint augmenting paths with respect to M
1
.
Proof. Consider the graph G

= (V, M
1
⊕ M
2
). Each vertex in V is
incident to at most one edge in M
2
−M
1
and at most one edge in M
1
−M
2
.
Thus, each connected component in G

is either
• an isolated vertex,
• a cycle of even length,
• a path of even length, or
• a path of odd length.
Moreover, the edges of all paths and cycles in G

are alternately in
M
2
−M
1
and M
1
−M
2
, which means that all cycles and even-length paths
have the same number of edges from M
1
as the number of edges from
M
2
. Since there are k more M
2
edges than M
1
edges in G

= (V, M
1
⊕M
2
) consists of an even length cycle, two isolated vertices and two
augmenting paths with respect to M
1
. On the other hand, [M
2
[ −[M
1
[ = 2.
/ o c
;
j
d
c o
/
| I ,
(b)
Matching in A
i
Matching in A
z
/ o c
;
j
d
c o
/
| I ,
(a)
Fig. 17.3 Illustration of Theorem 17.1. (a) M
2
. (b) M
1
⊕ M
2
.
17.3 The Network Flow Method
Recall that an undirected graph is called bipartite if it contains no cycles
of odd length. For example the graph in Fig. 17.1 is bipartite. Let G =
(X∪Y, E) be a bipartite graph. We can utilize one of the maximum network
ﬂow algorithms to ﬁnd a maximum matching in G as shown in Algorithm
bimatch1.
The correctness of the algorithm is easy to verify. It is also easy to
see that the construction of the ﬂow network takes no more than O(m)
time, where m = [E[. Its running time is dependent on the maximum ﬂow
algorithm used. If, for example, Algorithm mpm is used, then the running
time is O(n
3
), where n = [X[ +[Y [.
The Hungarian Tree Method for Bipartite Graphs 441
Algorithm 17.1 bimatch1
Input: A bipartite graph G = (X ∪ Y, E).
Output: A maximum matching M in G.
1. Direct all edges in G from X to Y .
2. Add a source vertex s and a directed edge (s, x) from s to x for each
vertex x ∈ X.
3. Add a sink vertex t and a directed edge (y, t) from y to t for each
vertex y ∈ Y .
4. Assign a capacity c(u, v) = 1 to each (directed) edge (u, v).
5. Use one of the maximum network ﬂow algorithms to ﬁnd a maxi-
mum ﬂow for the constructed network. M consists of those edges
connecting X to Y whose corresponding directed edge carries a ﬂow
of one unit.
17.4 The Hungarian Tree Method for Bipartite Graphs
Let G = (V, E) be an undirected graph. Lemma 17.1 and Corollary 17.1
suggest a procedure for ﬁnding a maximum matching in G. Starting from
an arbitrary (e.g. empty) matching, we ﬁnd an augmenting path p in G,
invert the roles of the edges in p (matched to unmatched and vice-versa),
and repeat the process until there are no more augmenting paths. At
that point, the matching, by Corollary 17.1, is maximum. Finding an
augmenting path in the case of bipartite graphs is much easier than in the
case of general graphs.
Let G = (X∪Y, E) be a bipartite graph with [X[+[Y [ = n and [E[ = m.
Let M be a matching in G. We call a vertex in X an x-vertex. Similarly
a y-vertex denotes a vertex in Y . First, we pick a free x-vertex, say r, and
label it outer. From r, we grow an alternating path tree, i.e., a tree in which
each path from the root r to a leaf is an alternating path. This tree, call
it T, is constructed as follows. Starting from r, add each unmatched edge
(r, y) connecting r to the y-vertex y and label y inner. For each y-vertex
y adjacent to r, add the matched edge (y, z) to T if such a matched edge
exists, and label z outer. Repeat the above procedure and extend the tree
until either a free y-vertex is encountered or the tree is blocked, i.e., cannot
be extended any more (note that no vertex is added to the tree more than
once). If a free y-vertex is found, say v, then the alternating path from the
root r to v is an augmenting path. On the other hand, if the tree is blocked,
then in this case the tree is called a Hungarian tree. Next, we start from
442 Matching
another free x-vertex, if any, and repeat the above procedure.
If T is a Hungarian tree, then it cannot be extended; each alternating
path traced from the root is stopped at some outer vertex. The only free
vertex in T is its root. Notice that if (x, y) is an edge such that x is in
T and y is not in T, then x must be labeled inner. Otherwise, x must be
connected to a free vertex or T is extendable through x. It follows that no
vertex in a Hungarian tree can occur in an augmenting path. For suppose
that p is an alternating path that